Appendix G Selected Solutions
2 Classification
2.4 Exercises
✍🏻 Problems
Exercises
2. Determine Whether the Differential Equation Is Linear.
4. Determine the Linearity of Each Term.
Solution.
To classify terms, focus on the dependent variable \(y\) and its derivatives. The table below summarizes the linearity of each term:
| Term | Coefficient | Dependent Variable | Linearity |
| \(e^t y^{(7)}\) | \(e^t\) | \(y^{(7)}\) | Linear |
| \((t+1)y'y'''\) | \((t+1)\) | \(y'y'''\) | Nonlinear |
| \(t \ln y''\) | \(t\) | \(\ln y''\) | Nonlinear |
| \(y' \sin t\) | \(\sin t\) | \(y'\) | Linear |
| \(\tan y\) | \(1\) | \(\tan y\) | Nonlinear |
| \(\dfrac{4}{y}\) | \(4\) | \(\dfrac{1}{y}\) | Nonlinear |
| \(\dfrac{3}{t}\) | \(-\) | \(-\) | Linear |
Note the last term, \(\dfrac{3}{t}\text{,}\) is a free term since it does not involve the dependent variable \(y\) or its derivatives. These terms are always linear in the context of differential equations.
3 Solutions
3.6 Exercises
✍🏻 Problems
Exercises
1.
Solution.
We need to substitute into the differential equation. To substitute for the left-hand side, we need the second derivative, so we begin by finding that.
\begin{align*}
y \amp = c_1 \sin x + c_2 \cos x \\
\Rightarrow \quad y' \amp = c_1 \cos x - c_2 \sin x \\
\Rightarrow \quad y'' \amp = -c_1 \sin x - c_2 \cos x
\end{align*}
Now we can substitute into the differential equation:
\begin{align*}
\frac{d^2y}{dx^2} + y \amp = 0 \\
\Big[ -c_1 \sin x - c_2 \cos x \Big] + \Big[ c_1 \sin x + c_2 \cos x \Big] \amp = 0 \\
-c_1 \sin x - c_2 \cos x + c_1 \sin x + c_2 \cos x \amp = 0 \\
0 \amp = 0 \quad
\end{align*}
Since substituting into the differential equation makes the LHS equal the RHS, we have verified that \(y = c_1 \sin x + c_2 \cos x\) is a solution to \(\ds \frac{d^2y}{dx^2} + y = 0.\)
2.
Solution.
The function \(\ds y = c_{1}e^{2x} + c_{2}xe^{2x}\) is a solution to the differential equation
\begin{equation*}
\frac{d^{2}y}{dx^{2}} - 4\frac{dy}{dx} + 4y = 0
\end{equation*}
since plugging it in gives
\begin{equation*}
4c_{1}e^{2x} + 4c_{2}e^{2x} + 4c_{2}xe^{2x} - 4(2c_{1}e^{2x} + c_{2}e^{2x} + 2c_{2}xe^{2x}) + 4(c_{1}e^{2x} + c_{2}xe^{2x}) = 0
\end{equation*}
\begin{equation*}
0 = 0 \quad
\end{equation*}
3.
Solution.
Find \(y'\) and \(y''\) since they appear in the equation and move all terms to the left side.
\begin{align*}
y \amp = 3\sin(x^2) \\
y' \amp = 3\cos(x^2) \cdot 2x = 6x\cos(x^2) \\
y'' \amp = 6\cos(x^2) - 12x^2\sin(x^2)
\end{align*}
Plug in \(y, y',\) and \(y''\) and simplify.
\begin{align*}
y' - xy'' - 12x^3\sin(x^2)
\amp = 0 \\
6x\cos(x^2) - x\left( 6\cos(x^2) - 12x^2\sin(x^2) \right) - 12x^3\sin(x^2)
\amp = 0 \\
\cancel{6x\cos(x^2)} - \cancel{6x\cos(x^2)} + \cancel{12x^3\sin(x^2)} - \cancel{12x^3\sin(x^2)}
\amp = 0\\
0 \amp = 0
\end{align*}
Therefore, \(y= 3\sin(x^2)\) is a solution to \(y' - xy'' = 12x^3\sin(x^2).\)
4.
Solution.
Working out the left and right-hand sides of this DE, we have:
\begin{align*}
LHS: \\
\frac{dy}{dx}
\amp = \frac{d}{dx}\left( Ce^{-5x^2} + \frac{3}{5} \right)\\
\amp = Ce^{-5x^2} \cdot \frac{d}{dx}(-5x^2) + 0\\
\amp = Ce^{-5x^2} \cdot -10x\\
\amp = -10Cxe^{-5x^2}
\end{align*}
\begin{align*}
RHS: \\
\big(2x\big)\amp\big(3 - 5y\big)\\
\amp = 2x \left[3 - 5\left(Ce^{-5x^2} + \frac{3}{5}\right) \right]\\
\amp = 2x \left[ 3 - 5Ce^{-5x^2} - 3 \right]\\
\amp = 2x \left[-5Ce^{-5x^2}\right]\\
\amp = -10Cxe^{-5x^2}
\end{align*}
Since \(LHS = RHS\text{,}\) we have verified that \(y = Ce^{-5x^2} + \frac{3}{5}\) is a solution to
\begin{equation*}
\frac{dy}{dx} = \big( 2x \big) \big( 3 - 5y \big)
\end{equation*}
5.
Solution.
First, we take derivatives to substitute into the differential equation.
\begin{align*}
y \amp = e^{rx} \\
\Rightarrow \quad y' \amp = re^{rx} \\
\Rightarrow \quad y'' \amp = r^2 e^{rx}
\end{align*}
Now we can substitute into the equation and solve for \(r\text{:}\)
\begin{align*}
\frac{d^2 y}{dx^2}
+ 6 \frac{dy}{dx} + 5y \amp = 0 \\
\Big[ r^2 e^{rx} \Big] + 6\Big[ r e^{rx} \Big]
+ 5\Big[ e^{rx} \Big] \amp = 0 \\
e^{rx} \Big[ r^2 + 6r + 5 \Big] \amp = 0 \\
e^{rx} (r+5)(r+1) \amp = 0
\end{align*}
Recall that \(e^{rx} \ne 0,\) no matter the value of \(x\) or \(r.\) Thus, we have:
\begin{align*}
r+5 = 0 \amp \text{ or } r+1 = 0 \\
r = -5, \amp \text{ or } r = -1
\end{align*}
So the two solutions are \(r = -5, -1.\)
6.
Solution.
First, we take derivatives to substitute into the differential equation.
\begin{align*}
y \amp = e^{rx} \\
\Rightarrow \quad y' \amp = re^{rx} \\
\Rightarrow \quad y'' \amp = r^2 e^{rx}
\end{align*}
Now we can substitute into the equation and solve for \(r\text{:}\)
\begin{align*}
y'' - 10y' + 24y \amp = 0 \\
\Big[ r^2 e^{rx} \Big] - 10\Big[ r e^{rx} \Big]
+ 24\Big[ e^{rx} \Big] \amp = 0 \\
e^{rx} \Big[ r^2 - 10r + 24 \Big] \amp = 0 \\
e^{rx} (r-4)(r-6) \amp = 0
\end{align*}
Recall that \(e^{rx} \ne 0,\) no matter the value of \(x\) or \(r.\) Thus, we have:
\begin{align*}
r-4 = 0 \amp \text{ or } r-6 = 0 \\
r = 4, \amp \text{ or } r = 6
\end{align*}
So the two solutions are \(r = 4, 6.\)
7.
Solution.
First, we take derivatives to substitute into the differential equation.
\begin{align*}
y \amp = e^{rx} \\
\Rightarrow \quad y' \amp = re^{rx} \\
\Rightarrow \quad y'' \amp = r^2 e^{rx}
\end{align*}
Now we can substitute into the equation and solve for \(r\text{:}\)
\begin{align*}
y'' - 25y \amp = 0 \\
\Big[ r^2 e^{rx} \Big] - 25\Big[ e^{rx} \Big] \amp = 0 \\
e^{rx} \Big[ r^2 - 25 \Big] \amp = 0 \\
e^{rx} (r-5)(r+5) \amp = 0
\end{align*}
Recall that \(e^{rx} \ne 0,\) no matter the value of \(x\) or \(r.\) Thus, we have:
\begin{align*}
r-5 = 0 \amp \text{ or } r+5 = 0 \\
r = 5, \amp \text{ or } r = -5
\end{align*}
So the two solutions are \(r = -5, 5.\)
8.
Solution.
First, we take derivatives to substitute into the differential equation.
\begin{align*}
y \amp = e^{rx} \\
\Rightarrow \quad y' \amp = re^{rx} \\
\Rightarrow \quad y'' \amp = r^2 e^{rx}
\end{align*}
Now we can substitute into the equation and solve for \(r\text{:}\)
\begin{align*}
7y'' \amp = 8y \\
7\Big[ r^2 e^{rx} \Big] \amp = 8\Big[ e^{rx} \Big] \\
7r^2 e^{rx} \amp = 8 e^{rx} \\
7r^2 e^{rx} - 8 e^{rx} \amp = 0 \\
e^{rx} \Big[ 7r^2 - 8 \Big] \amp = 0
\end{align*}
Recall that \(e^{rx} \ne 0,\) no matter the value of \(x\) or \(r.\) Thus, we have:
\begin{align*}
7r^2 - 8 \amp = 0 \\
r^2 \amp = \frac{8}{7} \\
r \amp = \pm\sqrt{\frac{8}{7}}
\end{align*}
So the two solutions are \(r = \pm\sqrt{\frac{8}{7}}.\)
9.
Solution.
We proceed as in the previous question. First, we take derivatives to substitute into the differential equation.
\begin{align*}
y \amp = t^k \\
\Rightarrow \quad y' \amp = kt^{k-1} \\
\Rightarrow \quad y'' \amp = k(k-1)t^{k-2}
\end{align*}
Now we can substitute into the differential equation:
\begin{align*}
3t^2 \frac{d^2y}{dt^2} \amp =
-11t \frac{dy}{dt} + 3y \\
3t^2\Big[ k(k-1)t^{k-2} \Big] \amp =
-11t\Big[ kt^{k-1} \Big] + 3\Big[ t^k \Big] \\
3k(k-1)t^2 t^{k-2} \amp =
-11kt^1\cdot t^{k-1} + 3t^k \\
(3k^2 - 3k)t^{2+(k-2)} \amp =
-11k t^{1 + (k-1)} + 3t^k \\
(3k^2 - 3k)t^{k} \amp = -11k t^{k} + 3t^k \\
(3k^2 - 3k)t^{k} + 11k t^{k} - 3t^k \amp = 0 \\
(3k^2 - 3k + 11k - 3)t^k \amp = 0 \\
(3k^2 +8k - 3)t^k \amp = 0 \\
(3k -1)(k+3)t^k \amp = 0
\end{align*}
Note that there is no value of \(k\) that can guarantee that \(t^k\) is 0. Hence we have
\begin{align*}
3k-1 = 0 \amp \quad\text{or}\quad k+3 = 0 \\
k = \frac{1}{3}, \amp \quad\text{or}\quad k = -3
\end{align*}
So the two solutions are \(k = -3, \frac{1}{3}.\)
10.
Solution.
We proceed as in the previous question. First, we take derivatives to substitute into the differential equation.
\begin{align*}
y \amp = t^k \\
\Rightarrow \quad y' \amp = kt^{k-1} \\
\Rightarrow \quad y'' \amp = k(k-1)t^{k-2}
\end{align*}
Now we can substitute into the differential equation:
\begin{align*}
t^{2}y''- 11ty' + 32y \amp = 0 \\
t^2\Big[ k(k-1)t^{k-2} \Big] - 11t\Big[ kt^{k-1} \Big] + 32\Big[ t^k \Big] \amp = 0 \\
k(k-1)t^2 t^{k-2} - 11kt^1\cdot t^{k-1} + 32t^k \amp = 0 \\
(k^2 - k)t^{2+(k-2)} - 11k t^{1 + (k-1)} + 32t^k \amp = 0 \\
(k^2 - k)t^{k} - 11k t^{k} + 32t^k \amp = 0 \\
(k^2 - k - 11k + 32)t^k \amp = 0 \\
(k^2 - 12k + 32)t^k \amp = 0 \\
(k-4)(k-8)t^k \amp = 0
\end{align*}
Note that there is no value of \(k\) that can guarantee that \(t^k\) is 0. Hence we have
\begin{align*}
k-4 = 0 \amp \quad\text{or}\quad k-8 = 0 \\
k = 4, \amp \quad\text{or}\quad k = 8
\end{align*}
So the two solutions are \(k = 4, 8.\)
11.
12.
Solution.
Given that \(y = \sin(mt)\text{,}\) we find the derivatives:
\begin{align*}
y \amp = \sin(mt) \\
y' \amp = m\cos(mt) \\
y'' \amp = -m^2\sin(mt)
\end{align*}
Now substitute into the differential equation:
\begin{align*}
y'' + 9y \amp = 0 \\
-m^2\sin(mt) + 9\sin(mt) \amp = 0 \\
(-m^2 + 9)\sin(mt) \amp = 0
\end{align*}
Since \(\sin(mt)\) is not always zero, we must have:
\begin{align*}
-m^2 + 9 \amp = 0 \\
m^2 \amp = 9 \\
m \amp = \pm 3
\end{align*}
Therefore, \(m = \pm 3\text{.}\)
13.
Solution.
Given that \(y = e^{3t}\text{,}\) we find the derivatives:
\begin{align*}
y \amp = e^{3t} \\
\frac{dy}{dt} \amp = 3e^{3t} \\
\frac{d^2y}{dt^2} \amp = 9e^{3t}
\end{align*}
Now substitute into the differential equation:
\begin{align*}
\frac{d^2y}{dt^2} - 8 \frac{dy}{dt} + my \amp = 0 \\
9e^{3t} - 8(3e^{3t}) + me^{3t} \amp = 0 \\
9e^{3t} - 24e^{3t} + me^{3t} \amp = 0 \\
(9 - 24 + m)e^{3t} \amp = 0 \\
(-15 + m)e^{3t} \amp = 0
\end{align*}
Since \(e^{3t} \ne 0\) for any value of \(t\text{,}\) we must have:
\begin{align*}
-15 + m \amp = 0 \\
m \amp = 15
\end{align*}
Therefore, \(m = 15\text{.}\)
14. Answer the following.
Solution.
-
To show this, let’s first move all terms to one side of the equation, so\begin{equation*} \os{\large{\text{LHS}}}{\overline{x^2y^\prime + xy - 1}} = 0 \end{equation*}Now, we will compute \(y^\prime\) and plug it and \(y\) into the left-hand side (LHS) of the equation to see if it simplifies to zero.\begin{align*} \os{\large{\text{LHS}}}{\overline{x^2y^\prime + xy - 1}} \amp = x^2\left(\frac{1 - \ln x - c}{x^2}\right) + x\left(\frac{\ln x + c}{x}\right) - 1\\ \amp = 1 - \ln x - c + \ln x + c - 1 \\ \amp = 0 \quad \end{align*}Applying the initial condition \(y(9)=8\) gives\begin{equation*} 8 = \frac{\ln 9 + c}{9} \quad \implies \quad c = 72-\ln(9) \end{equation*}and so the particular solution is\begin{equation*} \ds y = \frac{\ln x + 72-\ln(9)}{x} \end{equation*}
15. Verifying Particular Solutions of Initial-Value Problems Takes Two Steps.
Solution.
To verify that the function \(y = -3e^{x^2} + 3\) is a particular solution to the initial-value problem
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \qquad y(0) = 0,
\end{equation*}
we must show the following two things:
-
\(y = -3e^{x^2} + 3\) satisfies the differential equation.
-
\(y = -3e^{x^2} + 3\) satisfies the initial condition.
Let’s start by showing that the function \(y = -3e^{x^2} + 3\) satisfies the differential equation.
\begin{align*}
\frac{dy}{dx} \amp = 2xy - 6x \\
\frac{d}{dx}\left[-3e^{x^2} + 3\right] \amp = 2x(-3e^{x^2} + 3) - 6x \\
-3\frac{d}{dx}\left[e^{x^2}\right]+ 0 \amp = -6xe^{x^2} + 6x - 6x \\
-3\left(e^{x^2} \cdot 2x\right) \amp = -6xe^{x^2} \\
-6xe^{x^2} \amp = -6xe^{x^2}\quad
\end{align*}
Now, let’s show that the function satisfies the initial condition \(y(0) = 0\text{.}\)
\begin{align*}
y(0) \amp = -3e^{0^2} + 3 \\
0 \amp = -3e^0 + 3 \\
0 \amp = -3 + 3 \\
0 \amp = 0\quad
\end{align*}
16.
Solution.
We must show that \(y = 3x^2 - 2x + 1\) satisfies both the differential equation and the initial condition.
First, we check if \(y = 3x^2 - 2x + 1\) satisfies the differential equation.
\begin{align*}
\frac{dy}{dx} \amp = 6x - 2 \\
\frac{d}{dx}\left[3x^2 - 2x + 1\right] \amp = 6x - 2 \\
6x - 2 \amp = 6x - 2 \quad
\end{align*}
Now, we check if \(y = 3x^2 - 2x + 1\) satisfies the initial condition \(y(0) = 1\text{.}\)
\begin{align*}
y(0) \amp = 3(0)^2 - 2(0) + 1 \\
1 \amp = 1 \quad
\end{align*}
Since \(y = 3x^2 - 2x + 1\) satisfies both the differential equation and the initial condition, it is a particular solution to the initial-value problem.
17.
Solution.
To verify that the function \(y = -3e^{x^2} + 3\) is a particular solution to the initial-value problem
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \qquad y(0) = 0\text{,}
\end{equation*}
we must show the following two things:
-
\(y = -3e^{x^2} + 3\) satisfies the differential equation.
-
\(y = -3e^{x^2} + 3\) satisfies the initial condition.
Let’s start by showing that the function \(y = -3e^{x^2} + 3\) satisfies the differential equation.
\begin{align*}
\frac{dy}{dx} \amp = 2xy - 6x \\
\frac{d}{dx}\left[-3e^{x^2} + 3\right] \amp = 2x(-3e^{x^2} + 3) - 6x \\
-3\frac{d}{dx}\left[e^{x^2}\right]+ 0 \amp = -6xe^{x^2} + 6x - 6x \\
-3\left(e^{x^2} \cdot 2x\right) \amp = -6xe^{x^2} \\
-6xe^{x^2} \amp = -6xe^{x^2}\quad
\end{align*}
Now, let’s show that the function satisfies the initial condition \(y(0) = 0\text{.}\)
\begin{align*}
y(0) \amp = -3e^{0^2} + 3 \\
0 \amp = -3e^0 + 3 \\
0 \amp = -3 + 3 \\
0 \amp = 0\quad
\end{align*}
18. Find the Correct Solution.
Solution.
-
To show this, let’s first move all terms to one side of the equation, so\begin{equation*} \os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}} = 0 \end{equation*}For each of the functions above, we will compute \(y^\prime\) and plug it and \(y\) into the left-hand side (LHS) of the equation to see if it simplifies to zero.
-
\begin{equation*} y=c\sin(-x) \quad \implies \quad y^\prime = -c\cos(-x) \end{equation*}\begin{align*} \os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}} \amp = -c\cos(-x) - c\sin(-x) + c^2\sin^2(-x)\\ \amp = -c\cos(x) + c\sin(x) + c^2\sin^2(x) \end{align*}It can be difficult to see why this doesn’t simplify to zero. However, if we assume it does, then plugging any \(x\) into this expression must also give you zero. So, if you can find one \(x\) value where this doesn’t happen, then this function cannot be a solution.Letting \(x=0\) gives \(-c\text{,}\) but \(c\) can be any constant, so \(y = c\sin(-x)\) is not a solution.
-
\begin{equation*} y = \frac{1}{c+x^2} \quad \implies \quad y^\prime = \frac{-2x}{(c+x^2)^2} \end{equation*}\begin{align*} \os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}} \amp = \frac{-2x}{(c+x^2)^2} - \frac{1}{c+x^2} - \frac{1}{(c+x^2)^2}\\ \amp = \frac{-2x}{(c+x^2)^2} - \frac{1}{c+x^2}\cdot \frac{c+x^2}{c+x^2} - \frac{1}{(c+x^2)^2}\\ \amp = \frac{-2x}{(c+x^2)^2} - \frac{c+x^2}{(c+x^2)^2} - \frac{1}{(c+x^2)^2} \\ \amp = \frac{-2x-c-x^2-1}{(c+x^2)^2} \ne 0 \quad \end{align*}Letting \(x=0\) gives \(-1/c\text{,}\) but since \(c\) can be any constant, \(\ds y = \frac{1}{c+x^2}\) is not a solution.
-
\begin{equation*} y = \frac{1}{1+ce^{-x}} \quad \implies \quad y^\prime = \frac{ce^{-x}}{(1+ce^{-x})^2} \end{equation*}\begin{align*} \os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}} \amp = \frac{ce^{-x}}{(1+ce^{-x})^2} - \frac{1}{1+ce^{-x}} - \frac{1}{(1+ce^{-x})^2}\\ \amp = \frac{ce^{-x}}{(1+ce^{-x})^2} - \frac{1+ce^{-x}}{(1+ce^{-x})^2} - \frac{1}{(1+ce^{-x})^2}\\ \amp = \frac{ce^{-x}-1-ce^{-x}-1}{(1+ce^{-x})^2} \\ \amp = \frac{0}{(1+ce^{-x})^2} = 0 \quad \end{align*}So \(\ds y = \frac{1}{1+ce^{-x}}\) is a solution.
-
-
Setting \(y(0)=7\text{,}\) we get\begin{equation*} 7 = \frac{1}{1+c} \quad \implies \quad c = \frac{1}{7}-1 = -\frac{6}{7} \end{equation*}So the particular solution for this initial condition is\begin{equation*} y = \frac{1}{1-\frac{6}{7}e^{-x}} \end{equation*}
4 Direct Integration
4.3 Exercises
🏋️♂️ Practice Drills
Exercises
1. \(y''-9y = 0\).
Solution.
To verify if a function is a solution, we need to compute \(y''\) and check if \(y'' - 9y = 0\text{.}\)
-
\(y = 3\text{:}\) \(y'' = 0\text{,}\) so \(y'' - 9y = 0 - 27 = -27 \neq 0\text{.}\) Not a solution.
-
\(y = e^{3x}\text{:}\) \(y'' = 9e^{3x}\text{,}\) so \(y'' - 9y = 9e^{3x} - 9e^{3x} = 0\text{.}\) Solution.
-
\(y = 3x\text{:}\) \(y'' = 0\text{,}\) so \(y'' - 9y = 0 - 27x = -27x \neq 0\text{.}\) Not a solution.
-
\(y = 9e^x\text{:}\) \(y'' = 9e^x\text{,}\) so \(y'' - 9y = 9e^x - 81e^x = -72e^x \neq 0\text{.}\) Not a solution.
-
\(y = x^9\text{:}\) \(y'' = 72x^7\text{,}\) so \(y'' - 9y = 72x^7 - 9x^9 \neq 0\text{.}\) Not a solution.
-
\(y = 4e^{3x}\text{:}\) \(y'' = 36e^{3x}\text{,}\) so \(y'' - 9y = 36e^{3x} - 36e^{3x} = 0\text{.}\) Solution.
-
\(y = e^{-3x}\text{:}\) \(y'' = 9e^{-3x}\text{,}\) so \(y'' - 9y = 9e^{-3x} - 9e^{-3x} = 0\text{.}\) Solution.
-
\(y = e^{3x} - 2e^{-3x}\text{:}\) \(y'' = 9e^{3x} - 18e^{-3x}\text{,}\) so \(y'' - 9y = (9e^{3x} - 18e^{-3x}) - 9(e^{3x} - 2e^{-3x}) = 0\text{.}\) Solution.
2. \(y'' - 10y' + 25y = 0\).
Solution.
To verify if a function is a solution, we need to compute \(y'\) and \(y''\) and check if \(y'' - 10y' + 25y = 0\text{.}\)
-
\(y = 0\text{:}\) \(y' = 0\text{,}\) \(y'' = 0\text{,}\) so \(y'' - 10y' + 25y = 0 - 0 + 0 = 0\text{.}\) Solution.
-
\(y = x^2e^{5x}\text{:}\) \(y' = 2xe^{5x} + 5x^2e^{5x}\text{,}\) \(y'' = 2e^{5x} + 10xe^{5x} + 10xe^{5x} + 25x^2e^{5x} = (2 + 20x + 25x^2)e^{5x}\text{.}\) Then \(y'' - 10y' + 25y = (2 + 20x + 25x^2)e^{5x} - 10(2x + 5x^2)e^{5x} + 25x^2e^{5x} = 2e^{5x} \neq 0\text{.}\) Not a solution.
-
\(y = e^{5x}\text{:}\) \(y' = 5e^{5x}\text{,}\) \(y'' = 25e^{5x}\text{,}\) so \(y'' - 10y' + 25y = 25e^{5x} - 50e^{5x} + 25e^{5x} = 0\text{.}\) Solution.
-
\(y = 5x\text{:}\) \(y' = 5\text{,}\) \(y'' = 0\text{,}\) so \(y'' - 10y' + 25y = 0 - 50 + 125x \neq 0\text{.}\) Not a solution.
-
\(y = xe^{5x}\text{:}\) \(y' = e^{5x} + 5xe^{5x}\text{,}\) \(y'' = 5e^{5x} + 5e^{5x} + 25xe^{5x} = (10 + 25x)e^{5x}\text{,}\) so \(y'' - 10y' + 25y = (10 + 25x)e^{5x} - 10(1 + 5x)e^{5x} + 25xe^{5x} = 0\text{.}\) Solution.
-
\(y = 5e^{5x}\text{:}\) \(y' = 25e^{5x}\text{,}\) \(y'' = 125e^{5x}\text{,}\) so \(y'' - 10y' + 25y = 125e^{5x} - 250e^{5x} + 125e^{5x} = 0\text{.}\) Solution.
-
\(y = 1/25\text{:}\) \(y' = 0\text{,}\) \(y'' = 0\text{,}\) so \(y'' - 10y' + 25y = 0 - 0 + 1 = 1 \neq 0\text{.}\) Not a solution.
-
\(y = e^{-5x}\text{:}\) \(y' = -5e^{-5x}\text{,}\) \(y'' = 25e^{-5x}\text{,}\) so \(y'' - 10y' + 25y = 25e^{-5x} + 50e^{-5x} + 25e^{-5x} = 100e^{-5x} \neq 0\text{.}\) Not a solution.
-
\(y = (1+x)e^{5x}\text{:}\) \(y' = e^{5x} + 5(1+x)e^{5x} = (1 + 5 + 5x)e^{5x} = (6 + 5x)e^{5x}\text{,}\) \(y'' = 5e^{5x} + 5(6 + 5x)e^{5x} = (5 + 30 + 25x)e^{5x} = (35 + 25x)e^{5x}\text{.}\) Then \(y'' - 10y' + 25y = (35 + 25x)e^{5x} - 10(6 + 5x)e^{5x} + 25(1+x)e^{5x} = 0\text{.}\) Solution.
✍🏻 Problems
Exercises
16.
Solution.
Start by isolating the derivative, \(y'\text{,}\) on one side of the equation
\begin{align*}
y' \amp = 1 + 2 \sin x
\end{align*}
Integrate both sides with respect to \(x\) to leave us with \(y\) on the left side
\begin{align*}
\int y'\ dx \amp = \int \left(1 + 2 \sin x\right) \ dx \\
y + c_1 \amp = x - 2 \cos x + c_2 \\
y \amp = x - 2 \cos x + \boxed{c_2 - c_1}\leftarrow c \\
y(x) \amp = x - 2 \cos x + c
\end{align*}
Finally, apply the initial condition \(y(0) = 5\) to find \(c\)
\begin{align*}
y(0) \amp = 5 \\
(0) - 2 \cos(0) + c \amp = 5 \\
0 - 2 + c \amp = 5 \\
c \amp = 7
\end{align*}
Thus, the solution to the differential equation is \(\quad y = x - 2 \cos x + 7\text{.}\)
19. Solve the Equation.
Solution.
Start by isolating the derivative, \(y'\text{,}\) on one side of the equation
\begin{equation*}
y' = 1 + 2 \sin x \text{.}
\end{equation*}
Integrate both sides with respect to \(x\) to leave us with \(y\) on the left side
\begin{align*}
\int y'\ dx \amp = \int \left(1 + 2 \sin x\right) \ dx \\
y + c_1 \amp = x - 2 \cos x + c_2 \\
y \amp = x - 2 \cos x + \boxed{c_2 - c_1}\leftarrow c \\
y(x) \amp = x - 2 \cos x + c
\end{align*}
Finally, apply the initial condition \(y(0) = 5\) to find \(c\)
\begin{align*}
y(0) \amp = 5 \\
(0) - 2 \cos(0) + c \amp = 5 \\
0 - 2 + c \amp = 5 \\
c \amp = 7
\end{align*}
Thus, the solution to the differential equation is \(\quad y = x - 2 \cos x + 7\text{.}\)
Preview of a Future Method
1. Give the equation that can be rewritten in the form \(\ds\left[x^7 y \right]^{\prime} = e^x\).
5 Separation of Variables
5.5 Exercises
💡 Conceptual Quiz
Exercises
3. Short-Answer Questions.
3.e Thinking in Cases.
Solution.
Attempting to isolate the derivative on the left, we get:
\begin{equation*}
\frac{dy}{dx} = Q(x) - P(x) y
\end{equation*}
Since \(Q(x)\) and \(P(x)\) can only contain the variable \(x\) or constants, then this form is only separable in the following cases:
-
Both \(Q(x) = Q_0\) and \(P(x)=P_0\) are constant. Then\begin{equation*} \frac{dy}{dx} = Q_0 - P_0 y = (1) \cdot (Q_0 - P_0 y) \end{equation*}
-
\(Q(x) = 0\) and \(P(x)\) is anything. Then\begin{equation*} \frac{dy}{dx} = 0 - P(x) y = (-P(x)) \cdot (y) \end{equation*}
-
\(Q(x) = 0\) is anything and \(P(x) = 0\text{.}\) Then\begin{equation*} \frac{dy}{dx} = Q(x) - (0) y = (Q(x)) \cdot (1) \end{equation*}
🏋️♂️ Practice Drills
Exercises
1. Verify the solution.
Solution.
\begin{align*}
z' - 1 \amp = z^2\\
\frac{d}{dt}\Big( \tan(t+C) \Big) - 1 \amp = \Big( \tan(t+C) \Big)^2\\
\sec^2(t+C)\cdot \frac{d}{dt}(t+C) - 1 \amp = \tan^2(t+C)\\
\sec^2(t+C)\cdot (1+0) - 1 \amp = \sec^2(t+C) - 1\\
\sec^2(t+C) - 1 \amp = \tan^2(t+C) \quad
\end{align*}
Now, let’s find \(C\) with the initial condition \(z(4) = 9\text{:}\)
\begin{equation*}
9 = z(4) = \tan(4+C)\quad\Rightarrow\quad C = \arctan(9) - 4
\end{equation*}
2. Show the differential equation is separable.
Solution.
To see that this equation is separable, we need to check that it can be rearranged in separable form. Isolating \(s' \) we get
\begin{equation*}
s' = \frac{s + t}{s}
\end{equation*}
and it should be clear that there is no way to separate \(t\) and \(s\) by multiplication in the numerator, so this equation is not separable.
3.
4.
Solution.
This doesn’t look separable; but if we use a property of logarithms we can re-write as follows.
\begin{align*}
s' \amp = t\ln(s^{2t}) + 8t^2 \\
\frac{ds}{dt} \amp = t \cdot 2t\ln(s) + 8t^2 \\
\amp = 2t^2 \ln(s) + 8t^2 \\
\amp = \big(t^2\big)\big(2\ln(s) + 8\big)
\end{align*}
Hence, this DE is separable.
5.
6.
7.
8.
✍🏻 Problems
Exercises
2. Reorder the Steps.
Solution.
The correct order of steps for solving this separable differential equation is:
-
Check for 1st-order. \(\checkmark\) (The equation contains only \(\frac{dy}{dx}\text{,}\) so it is first-order.)
-
\(\dfrac{1}{y} dy = (1 + \cos(x))dx\) (Separate variables.)
-
\(\ds\int \frac{1}{y} dy = \int (1 + \cos(x))dx\) (Set up integrals.)
-
\(\ln|y| = x + \sin(x) + C\) (Integrate both sides and combine constants.)
3. Step-By-Step Initial-Valued Problem.
Solution.
We solve the initial value problem \(\dfrac{du}{dy} - u^2y = u^2\text{,}\) \(u(0) = 5\) step by step:
-
The equation is first-order since it contains only the first derivative \(\dfrac{du}{dy}\text{.}\)
-
Rearranging the equation:\begin{align*} \frac{du}{dy} \amp = u^2y + u^2 \\ \frac{du}{dy} \amp = u^2(y + 1) = u^2(1 + y) \end{align*}This is the separable form.
-
Integrating both sides:\begin{align*} \int \frac{1}{u^2}du \amp = \int (y+1)dy \\ -\frac{1}{u} \amp = \frac{y^2}{2} + y + c \end{align*}
-
Solving for \(u\text{:}\)\begin{equation*} u = \left(-\frac{y^2}{2} - y - c\right)^{-1} \end{equation*}
-
Using the initial condition \(u(0) = 5\text{:}\)\begin{align*} 5 \amp = \left(-\frac{0^2}{2} - 0 - c\right)^{-1} = (-c)^{-1} = -\frac{1}{c} \\ c \amp = -\frac{1}{5} \end{align*}Therefore, the particular solution is:\begin{equation*} u = \left(-\frac{y^2}{2} - y + \frac{1}{5}\right)^{-1} \end{equation*}
8.
Solution.
This equation is not obviously separable as written, so we begin by solving for \(y'\text{:}\)
\begin{align*}
y^2 y' \amp = 6x^2 + 1 \\
\frac{dy}{dx} \amp = \frac{6x^2 + 1}{y^2} = \big( 6x^2 + 1 \big)\cdot \left( \frac{1}{y^2} \right)
\end{align*}
This is now in separable form.
We now separate and integrate:
\begin{align*}
y^2 dy \amp = (6x^2 + 1) dx \\
\int y^2 dy \amp = \int (6x^2 + 1) dx \\
\frac{y^3}{3} \amp = 2x^3 + x + c
\end{align*}
Multiplying through and isolating \(y\) gives the general solution:
\begin{equation*}
y = \sqrt[3]{6x^3 + 3x + c}
\end{equation*}
18.
19.
20.
Solution.
\begin{align*}
\frac{dy}{dx} \amp = \frac{1 - x^2}{y^2} \\
y^2 dy \amp = (1-x^2)dx \\
\int y^2 dy \amp = \int (1-x^2)dx \\
\frac{1}{3}y^3 \amp = x - \frac{1}{3}x^3 + C_1 \\
y^3 \amp = 3x - x^3 + 3C_1 \\
y \amp = \sqrt[3]{3x - x^3 + 3C_1}, \mbox{ or} \\
y \amp = \sqrt[3]{3x - x^3 + C_2}
\end{align*}
Note: since, in general \(\ds \sqrt[3]{a+b} \ne \sqrt[3]{a} + \sqrt[3]{b} \) , we CANNOT simplify the solution to \(\ds y = \sqrt[3]{3x - x^3} + C.\)
21.
Solution.
\begin{align*}
y' - 2y \amp = y\sin x \\
y' \amp = 2y + y\sin x \\
\frac{dy}{dx} \amp = y(2 + \sin x) \\
\frac{1}{y} dy \amp = (2 + \sin x)dx \\
\int \frac{1}{y} dy \amp = \int (2 + \sin x)dx \\
\ln|y| \amp = 2x - \cos x + C_1 \\
e^{\ln|y|} \amp = e^{2x - \cos x + C_1} \\
|y| \amp = e^{2x - \cos x}\cdot e^{C_1} \\
y \amp = \pm e^{2x - \cos x}\cdot e^{C_1} \\
y \amp = \underbrace{\pm e^{C_1}}_{\scriptsize = C_2}\cdot e^{2x - \cos x} \\
y \amp = C_2e^{2x - \cos x}
\end{align*}
22.
Solution.
\begin{align*}
x\frac{dv}{dx} \amp = \frac{1-4v^2}{3v} \\
\frac{dv}{dx} \amp = \frac{1-4v^2}{3v}\cdot \frac{1}{x} \\
\frac{3v}{1-4v^2} dv \amp = \frac{1}{x} dx \\
\int \frac{3v}{1-4v^2} dv \amp = \int \frac{1}{x} dx
\end{align*}
For the integral on the left-hand side, we will need to do a substitution. If we let \(\ds u = 1-4v^2, \) then \(\ds du = -8vdv, \) and therefore \(\ds -\frac{1}{8}du = vdv. \) Then we have the following:
\begin{align*}
3\int \frac{v}{1-4v^2} dv \amp = \int \frac{1}{x} dx \\
3\int \underbrace{\frac{1}{1-4v^2}}_{\scriptsize 1/u} \underbrace{vdv}_{\scriptsize -\frac{1}{8}du} \amp = \int \frac{1}{x} dx \\
3\int \frac{1}{u}\cdot -\frac{1}{8}du \amp = \int \frac{1}{x} dx \\
-\frac{3}{8} \int \frac{1}{u}du \amp = \int \frac{1}{x} dx \\
-\frac{3}{8} \ln|u| \amp = \ln|x| + C_1 \\
-\frac{3}{8} \ln|1 - 4v^2| \amp = \ln|x| + C_1 \\
\ln|1 - 4v^2| \amp = -\frac{8}{3} \left(\ln|x| + C_1\right) \\
\ln|1 - 4v^2| \amp = -\frac{8}{3}\ln|x| - \frac{8}{3}C_1 \\
\ln|1 - 4v^2| \amp = \ln\left(|x|^{-8/3}\right) - \frac{8}{3}C_1 \\
e^{\ln|1 - 4v^2|} \amp = e^{\ln\left(|x|^{-8/3}\right) - \frac{8}{3}C_1} \\
|1 - 4v^2| \amp = e^{\ln\left(|x|^{-8/3}\right)}\cdot e^{-\frac{8}{3}C_1} \\
|1 - 4v^2| \amp = |x|^{-8/3}\cdot e^{-\frac{8}{3}C_1} \\
1 - 4v^2 \amp = \pm e^{-\frac{8}{3}C_1}\cdot |x|^{-8/3} \\
1 - 4v^2 \amp = \underbrace{\pm e^{-\frac{8}{3}C_1}}_{\scriptsize = C_2}\cdot x^{-8/3} \\
1 - 4v^2 \amp = C_2x^{-8/3}
\end{align*}
\begin{align*}
- 4v^2 \amp = C_2x^{-8/3} - 1 \\
v^2 \amp = \frac{C_2}{-4}x^{-8/3} + \frac{1}{4} \\
v^2 \amp = \underbrace{\frac{C_2}{-4}}_{\scriptsize = C_3}x^{-8/3} + \frac{1}{4} \\
v^2 \amp = C_3x^{-8/3} + \frac{1}{4} \\
v \amp = \pm \sqrt{C_3x^{-8/3} + \frac{1}{4}}
\end{align*}
23.
Solution.
24.
25.
Solution.
Informally, we divide both side of the differential equation by \(y\) and multiply each side by \(dx\) to get
\begin{equation*}
\frac{dy}y = -6xdx.
\end{equation*}
\begin{equation*}
\int \frac{dy}y = \int (-6x)dx
\end{equation*}
\begin{equation*}
\ln|y|=-3x^2+C
\end{equation*}
\begin{equation*}
\ln|y|=-3x^2+C
\end{equation*}
\begin{equation*}
|y|=e^{-3x^2+C}
\end{equation*}
\begin{equation*}
y=\pm e^C e^{-3x^2}
\end{equation*}
and hence
\begin{equation*}
y(x) = Ae^{-3x^2} \qquad (A = \pm e^C)
\end{equation*}
The condition \(y(0) = 7\) yields \(A =7\text{,}\) so the desired solution is
\begin{equation*}
y(x) = 7e^{-3x^2}
\end{equation*}
26.
Solution.
First we find the general solution:
\begin{align*}
\frac{dz}{dt} \amp = 2tz^2 \\
\frac{1}{z^2}dz \amp = 2tdt \\
\int \frac{1}{z^2}dz \amp = \int 2tdt \\
-\frac{1}{z} \amp = t^2 + C_1 \\
z \amp = -\frac{1}{t^2 + C_1}
\end{align*}
Using the initial condition \(z(1) = 2\text{:}\)
\begin{align*}
2 \amp = -\frac{1}{1^2 + C_1} \\
2 \amp = -\frac{1}{1 + C_1} \\
C_1 \amp = -\frac{1}{2} - 1 \\
C_1 \amp = -\frac{3}{2}
\end{align*}
So the solution to the initial value problem is:
\begin{align*}
z(t) \amp = -\frac{1}{t^2 - \frac{3}{2}}
\end{align*}
27.
Solution.
First we find the general solution:
\begin{align*}
\frac{dy}{d\theta} \amp = y\sin\theta \\
\frac{1}{y}dy \amp = \sin\theta d\theta \\
\int \frac{1}{y}dy \amp = \int \sin\theta d\theta \\
\ln|y| \amp = -\cos\theta + C_1 \\
e^{\ln|y|} \amp = e^{-\cos\theta + C_1} \\
|y| \amp = e^{-\cos\theta}\cdot e^{C_1} \\
y \amp = \pm e^{C_1}e^{-\cos\theta} \\
\amp = \underbrace{\pm e^{C_1}}_{\scriptsize = C_2}e^{-\cos\theta} \\
\amp = C_2e^{-\cos\theta}
\end{align*}
Then we use the initial condition \(\ds y(\pi) = -3 \text{.}\)
\begin{align*}
-3 \amp = y(\pi) \\
\amp = C_2e^{-\cos\pi} \\
\amp = C_2e^{-(-1)} \\
\amp = C_2e \\
-\frac{3}{e} \amp = C_2, \mbox{ or} \\
C_2 \amp = -3e^{-1}
\end{align*}
Hence, the solution is \(\ds y = -3e^{-1}e^{-\cos\theta} \) , or \(\ds y = -3e^{-\cos\theta-1} \) .
28.
Solution.
First we find the general solution:
\begin{align*}
\frac{dy}{dx} - 8x^3e^{-2y} \amp = 0 \\
\frac{dy}{dx} \amp = 8x^3e^{-2y} \\
\frac{1}{e^{-2y}}dy \amp = 8x^3dx \\
e^{2y}dy \amp = 8x^3dx \\
\int e^{2y}dy \amp = \int 8x^3dx \\
\frac{1}{2}e^{2y} \amp = 2x^4 + C_1 \\
e^{2y} \amp = 4x^4 + 2C_1 \\
\ln(e^{2y}) \amp = \ln(4x^4 + 2C_1) \\
2y \amp = \ln(4x^4 + 2C_1) \\
y \amp = \frac{1}{2}\ln(4x^4 + 2C_1), \mbox{ or} \\
\amp = \frac{1}{2}\ln(4x^4 + C_2), \mbox{ or} \\
\amp = \ln\Big[(4x^4 + C_2)^{1/2}\Big], \mbox{ or} \\
\amp = \ln\Big[\sqrt{4x^4 + C_2}\Big]
\end{align*}
Then we use the initial condition \(\ds y(1) = 0 \text{.}\)
\begin{align*}
0 \amp = y(1) \\
0 \amp = \ln\Big[\sqrt{4\cdot 1^4 + C_2}\Big] \\
0 \amp = \ln\Big[\sqrt{4 + C_2}\Big] \\
e^0 \amp = e^{\ln\Big[\sqrt{4 + C_2}\Big]} \\
1 \amp = \sqrt{4 + C_2} \\
1^2 \amp = \Big[\sqrt{4 + C_2}\Big]^2 \\
1 \amp = 4+C_2 \\
-3 \amp = C_2
\end{align*}
Hence, the solution is \(\ds y = \frac{1}{2}\ln(4x^4 - 3) \) , or \(\ds y = \ln\Big[\sqrt{4x^4 - 3}\Big] \text{.}\)
29.
Solution.
We want to find a non-zero function \(\mu(x)\) such that
\begin{equation*}
\frac{d}{dx}\left[\mu\sin x\right] = \mu \cos x + 6 \mu \sin x.
\end{equation*}
Applying the product rule to the left side of this equation gives
\begin{equation*}
\mu \cos x + \frac{d\mu}{dx} \sin x = \mu \cos x + 6 \mu \sin x.
\end{equation*}
Subtracting \(\mu \cos x\) from both sides gives
\begin{equation*}
\frac{d\mu}{dx} \sin x = 6 \mu \sin x\text{,}
\end{equation*}
which implies that
\begin{equation*}
\frac{d\mu}{dx} = 6 \mu.
\end{equation*}
This is a separable differential equation that we can solve with separation of variables:
\begin{align*}
\frac{d\mu}{\mu} \amp = 6 dx \\
\int \frac{1}{\mu} d\mu \amp = \int 6 dx \\
\ln|\mu| \amp = 6x + c_1 \\
|\mu| \amp = e^{6x + c_1} = e^{c_1} e^{6x} \\
\mu \amp = \pm e^{c_1} e^{6x}
\end{align*}
But, \(\pm e^{c_1}\) is just a constant, so we can write this as
\begin{equation*}
\mu = ce^{6x}.
\end{equation*}
Since the question asks for a non-zero function, we can take \(c = 1\text{.}\) Thus, \(\mu = e^{6x}\) satisfies the given differential equation.
30.
Solution.
We want to find a non-zero function \(\mu(x)\) such that
\begin{equation*}
\frac{d}{dx}\left[\mu\ln x\right] = \frac{\mu}{x} + 2x \mu \ln x.
\end{equation*}
Applying the product rule to the left side of this equation gives
\begin{equation*}
\mu \frac{1}{x} + \frac{d\mu}{dx} \ln x = \frac{\mu}{x} + 2x \mu \ln x.
\end{equation*}
Subtracting \(\frac{\mu}{x}\) from both sides gives
\begin{equation*}
\frac{d\mu}{dx} \ln x = 2x \mu \ln x\text{,}
\end{equation*}
which implies that
\begin{equation*}
\frac{d\mu}{dx} = 2x \mu.
\end{equation*}
This is a separable differential equation that we can solve with separation of variables:
\begin{align*}
\frac{d\mu}{\mu} \amp = 2x\ dx \\
\int \frac{1}{\mu} d\mu \amp = \int 2x\ dx \\
\ln|\mu| \amp = x^2 + c_1 \\
|\mu| \amp = e^{x^2 + c_1} = e^{c_1} e^{x^2} \\
\mu \amp = \pm e^{c_1} e^{x^2}
\end{align*}
But, \(\pm e^{c_1}\) is just a constant, so we can write this as
\begin{equation*}
\mu = ce^{x^2}.
\end{equation*}
Since the question asks for a non-zero function, we can take \(c = 1\text{.}\) Thus, \(\mu = e^{x^2}\) satisfies the given differential equation.
31.
Solution.
We want to find a non-zero function \(\mu(x)\) such that
\begin{equation*}
\frac{d}{dx}\left[\mu y\right] = \mu \frac{dy}{dx} + 5 \mu y.
\end{equation*}
Applying the product rule to the left side of this equation gives
\begin{equation*}
\mu \frac{dy}{dx} + \frac{d\mu}{dx} y = \mu \frac{dy}{dx} + 5 \mu y.
\end{equation*}
Subtracting \(\mu \frac{dy}{dx}\) from both sides gives
\begin{equation*}
\frac{d\mu}{dx} y = 5 \mu y\text{,}
\end{equation*}
which implies that
\begin{equation*}
\frac{d\mu}{dx} = 5 \mu.
\end{equation*}
This is a separable differential equation that we can solve with separation of variables:
\begin{align*}
\frac{d\mu}{\mu} \amp = 5 dx \\
\int \frac{1}{\mu} d\mu \amp = \int 5 dx \\
\ln|\mu| \amp = 5x + c_1 \\
|\mu| \amp = e^{5x + c_1} = e^{c_1} e^{5x} \\
\mu \amp = \pm e^{c_1} e^{5x}
\end{align*}
But, \(\pm e^{c_1}\) is just a constant, so we can write this as
\begin{equation*}
\mu = ce^{5x}.
\end{equation*}
Since the question asks for a non-zero function, we can take \(c = 1\text{.}\) Thus, \(\mu = e^{5x}\) satisfies the given differential equation.
32.
Solution.
We take time \(t\) in minutes, with \(t = 0\) corresponding to 5:00 PM. We also assume (somewhat unrealistically) that at any instant the temperature \(T(t)\) of the roast is uniform throughout. We have \(T(t) \lt A = 375\text{,}\) \(T(0)=50\text{,}\) and \(T(75) =125\text{.}\) Hence
\begin{align*}
\ds \frac{dT}{dt} = k(375-T)\\
\int \frac{1}{375-T}dT\amp = \ds \int kdt\\
-\ln(375-T)=\amp kt+C
\end{align*}
Now \(T(0)=50\text{,}\) implies that \(B = 325\text{,}\) so \(T(t) = 375-325e^{-kt}\text{.}\) We also know that \(T = 125\) when \(t = 75\text{.}\) Substitution of these values in the preceding equation yields
\begin{equation*}
k=-\frac{1}{75}\ln\left(\frac{250}{325}\right)\approx 0.0035.
\end{equation*}
Hence we finally solve the equation
\begin{equation*}
150=375-325e^{-0.0035t}
\end{equation*}
for \(t = -[\ln(225/325)]/[0.0035] \approx 105\) (min), the total cooking time required. Because the roast was placed in the oven at 5:00 PM, it should be removed at about 6:45 PM.
6 Integrating Factor
6.1 Exercises
💡 Conceptual Quiz
6.1.1.
6.1.1.l Equations Outside Separation and IF Methods.
🏋️♂️ Practice Drills
6.1.1. Identifying Equations for the IF Method.
Solution.
The integrating factor method applies to first-order linear differential equations of the form \(y' + P(x)y = Q(x)\text{.}\) We check each equation:
-
\(y' + 2y = 3x\) — First-order linear. Can use IF.
-
\(\dfrac{y'}{t^2} + y = 17t\) — Multiply by \(t^2\) to get \(y' + t^2y = 17t^3\text{.}\) First-order linear. Can use IF.
-
\(y'' + 2x y = 0\) — Second-order equation (has \(y''\)). Cannot use IF.
-
\(y' + y^2 = 17t\) — Has \(y^2\) term, which makes it nonlinear. Cannot use IF.
-
\(\dfrac{y'}{t} + y = 17t\) — Multiply by \(t\) to get \(y' + ty = 17t^2\text{.}\) First-order linear. Can use IF.
-
\(\cos x \ y = y' - e^x\) — Rearrange as \(y' - \cos x \cdot y = e^x\text{.}\) First-order linear. Can use IF.
Four equations can be solved using the integrating factor method.
6.1.2.
Solution.
This DE is first-order and linear, so it can be solved using an integrating factor.
It is not separable. After isolating the derivative:
\begin{equation*}
\dfrac{dy}{dx} = \dfrac{y - \cos x}{x^2},
\end{equation*}
we see that \(y\) and \(x\) cannot be separated by multiplication or division.
Therefore, this DE can be solved using an integrating factor, but not by separating variables.
6.1.3.
Solution.
This DE is first-order but not linear because the dependent variable \(x\) appears inside the nonlinear term \(e^x\text{.}\)
It is also not separable. After isolating the derivative:
\begin{equation*}
\frac{dx}{dt} = e^x - xt,
\end{equation*}
there’s no way to write the right-hand side as a product of a function of \(x\) and a function of \(t\text{.}\)
Therefore, this DE cannot be solved using either technique.
6.1.4.
Solution.
This DE is first-order and linear, so it can be solved using an integrating factor.
It is also separable. Solving for the derivative:
\begin{align*}
(t^2 + 1) \frac{dy}{dt} \amp = yt - y\\
\frac{dy}{dt} \amp = \frac{yt - y}{t^2+1}\\
\frac{dy}{dt} \amp = \frac{y(t - 1)}{t^2+1}\\
\frac{dy}{dt} \amp = y \cdot \frac{t - 1}{t^2+1}
\end{align*}
Here, we can separate the variables: \(\dfrac{dy}{y} = \dfrac{t - 1}{t^2 + 1} dt\text{.}\)
Therefore, this DE can be solved both by using an integrating factor and by separating variables.
6.1.5.
Solution.
This DE is first-order but not linear because the dependent variable \(y\) is raised to a power (\(y^2\)).
It is separable. Solving for the derivative:
\begin{align*}
\frac{dy}{dt} - y^2 t \amp = t\\
\frac{dy}{dt} \amp = t + y^2 t\\
\frac{dy}{dt} \amp = t \cdot (1 + y^2)
\end{align*}
allows us to separate variables: \(\dfrac{dy}{1+y^2} = t \, dt\text{.}\)
Therefore, this DE can be solved by separating variables, but not using an integrating factor.
6.1.6.
Solution.
This DE is first-order and linear, so it can be solved using an integrating factor.
It is not separable. Solving for the derivative:
\begin{equation*}
\frac{dr}{d\theta} = 3r + \theta^3,
\end{equation*}
shows no way to separate \(r\) and \(\theta\) into separate multiplicative terms.
Therefore, this DE can be solved using an integrating factor, but not by separating variables.
✍🏻 Problems
6.1.1.
Solution.
Standard form: \(y' + (-4)y = x\text{,}\) so \(P(x) = -4\text{.}\) Integrating factor: \(\mu(x) = e^{\int -4 dx} = e^{-4x}\text{.}\)
\begin{align*}
e^{-4x}y' - 4e^{-4x}y \amp = xe^{-4x}\\
\frac{d}{dx}[e^{-4x}y] \amp = xe^{-4x}
\end{align*}
Integrate:
\begin{align*}
e^{-4x}y \amp = \int xe^{-4x} dx = -\frac{1}{4}xe^{-4x} - \frac{1}{16}e^{-4x} + C\\
y \amp = -\frac{1}{4}x - \frac{1}{16} + Ce^{4x}
\end{align*}
6.1.2.
Solution 1.
Solution 2. (detailed)
We put the equation in standard form to identify \(P(x)\) and compute the integrating factor.
\begin{equation*}
y' + \us{P}{\boxed{-\frac{1}{x^2}}}\ y = \frac{1}{x^2} \quad \implies \quad
\mu(x) = e^{\large \int -\sfrac{1}{x^2} dx} = e^{1/x}
\end{equation*}
Multiply by \(e^{1/x}\) and reverse the product rule on the left side.
\begin{align*}
e^{1/x}y' - \frac{e^{1/x}}{x^2} y \amp = \frac{e^{1/x}}{x^2} \\
\frac{d}{dx}\left[e^{1/x} y\right] \amp = \frac{e^{1/x}}{x^2}
\end{align*}
Next, integrate both sides and use \(u\)-substitution on the right side.
Finally, solve for \(y\text{:}\)
\begin{align*}
\int \frac{d}{dx}\left[e^{1/x} y\right] dx
\amp = \int \frac{e^{1/x}}{x^2} dx \qquad \rightarrow\\
e^{1/x} y \amp = \int -e^u du \\
e^{1/x} y \amp = -e^u + c \\
e^{1/x} y \amp = -e^{1/x} + c
\end{align*}
\begin{align*}
u \amp = \frac{1}{x} \\
du \amp = -\frac{1}{x^2} dx
\end{align*}
\begin{gather*}
y = -1 + ce^{-1/x}
\end{gather*}
6.1.3.
6.1.4.
Solution.
6.1.5.
6.1.6.
Solution 1.
Rewrite: \(M' - \frac{4}{t}M = t^6 + t\) with \(M' = \frac{dM}{dt}\text{.}\) Integrating factor: \(\mu(t) = e^{\int -\frac{4}{t} dt} = t^{-4}\text{.}\)
\begin{align*}
\frac{d}{dt}[t^{-4}M] \amp = t^2 + t^{-3}\\
t^{-4}M \amp = \frac{1}{3}t^{3} - \frac{1}{2}t^{-2} + C\\
M \amp = \frac{1}{3}t^{7} - \frac{1}{2}t^{2} + Ct^{4}
\end{align*}
Solution 2. (detailed)
As always, start by putting the equation in standard form:
\begin{align*}
\frac{1}{t} \frac{dM}{dt} - \frac{4}{t^2} \cdot M \amp = t^5 + 1 \\
\frac{dM}{dt} + \us{P}{\boxed{-\frac{4}{t}}} \cdot M \amp = t^6 + t \text{.}
\end{align*}
6.1.7.
6.1.8.
Solution.
Standard form: \(r' + \tan\theta r = \sec\theta\) with \(r' = \frac{dr}{d\theta}\text{.}\) Integrating factor: \(\mu(\theta) = e^{\int \tan\theta d\theta} = \sec\theta\text{.}\)
\begin{align*}
\frac{d}{d\theta}[\sec\theta r] \amp = \sec^2\theta\\
\sec\theta r \amp = \tan\theta + C\\
r \amp = \sin\theta + C\cos\theta
\end{align*}
6.1.9.
Solution.
Standard form: \(y' - \frac{1}{x} y = x e^x\text{.}\) Integrating factor: \(\mu(x) = e^{\int -\frac{1}{x} dx} = x^{-1}\text{.}\)
\begin{align*}
\frac{d}{dx}\left[\frac{1}{x}y\right] \amp = e^x\\
\frac{1}{x}y \amp = \int e^x dx = e^x + C\\
y \amp = x(e^x + C)
\end{align*}
Apply \(y(1) = e - 1\text{:}\)
\begin{equation*}
e - 1 = 1(e + C) \Rightarrow C = -1.
\end{equation*}
6.1.10.
Solution.
Divide by \(e^t\text{:}\) \(z' + 4z = e^{-t}\text{.}\) Integrating factor: \(\mu(t) = e^{\int 4 dt} = e^{4t}\text{.}\)
\begin{align*}
\frac{d}{dt}[e^{4t}z] \amp = e^{4t} e^{-t} = e^{3t}\\
e^{4t}z \amp = \frac{1}{3} e^{3t} + C\\
z \amp = \frac{1}{3}e^{-t} + Ce^{-4t}
\end{align*}
Apply \(z(0) = \frac{4}{3}\text{:}\)
\begin{equation*}
\frac{4}{3} = \frac{1}{3} + C \Rightarrow C = 1.
\end{equation*}
6.1.11.
Solution.
Standard form with \(P(x) = -3\text{.}\) Integrating factor: \(\mu(x) = e^{\int -3 dx} = e^{-3x}\text{.}\)
\begin{align*}
\frac{d}{dx}[e^{-3x}y] \amp = 6e^{-3x}\\
e^{-3x}y \amp = -2e^{-3x} + C\\
y \amp = -2 + Ce^{3x}
\end{align*}
7 Qualitative Methods
7.7 Exercises
✍🏻 Problems
7.7.1. Practice Problems.
7.7.2. Exploring Nonlinear Dynamics.
7.7.3. Exploring Chaos in Autonomous Systems.
7.7.4. Exploring Real-World Applications.
7.7.5. Find the Equilibrium Solution.
Solution.
To find equilibrium solutions, we set \(f(y) = y^2 = 0\text{.}\) This gives us the solution \(y = 0\text{.}\) So, \(y = 0\) is the only equilibrium solution for this equation.
This means that if the solution starts at \(y(0) = 0\text{,}\) it will remain at zero for all time. If it starts with \(y(0) > 0\text{,}\) it increases; if it starts with \(y(0) < 0\text{,}\) it increases toward \(0\) (remaining negative).
7.7.6. Finding Equilibrium Points.
Solution.
To find the equilibrium solutions, we set the right-hand side equal to zero:
\begin{equation*}
y^2 - y^4 = 0.
\end{equation*}
This has solutions:
-
\(\displaystyle y = 0\)
-
\(\displaystyle y = 1\)
-
\(\displaystyle y = -1\)
These are the equilibrium solutions where the slope is zero. At these points, the solution remains constant over time.
7.7.7. Phase Line Sketching.
Solution.
First, find equilibria by setting \(y^2 - 4y + 3 = 0\text{,}\) which factors as \((y-1)(y-3) = 0\text{,}\) giving \(y = 1\) and \(y = 3\text{.}\)
Test regions: For \(y < 1\text{,}\) try \(y=0\text{:}\) \(f(0) = 3 > 0\) (upward arrow). For \(1 < y < 3\text{,}\) try \(y=2\text{:}\) \(f(2) = 4 - 8 + 3 = -1 < 0\) (downward arrow). For \(y > 3\text{,}\) try \(y=4\text{:}\) \(f(4) = 16 - 16 + 3 = 3 > 0\) (upward arrow).
Phase line: Place dots at \(y=1\) and \(y=3\text{.}\) Draw upward arrows in regions \(y < 1\) and \(y > 3\text{,}\) and a downward arrow in region \(1 < y < 3\text{.}\)
Classification: \(y = 1\) is a sink (stable) since arrows point toward it from both sides. \(y = 3\) is a source (unstable) since arrows point away from it on both sides.
8 Euler’s Method
8.4 Exercises
✍🏻 Problems
8.4.1.
Solution.
Using Euler’s method with \(h = 0.5\text{:}\)
-
\(x_0 = 0\text{,}\) \(y_0 = -1\)
-
\(\displaystyle y_1 = y_0 + 0.5 \cdot (y_0^2 - x_0) = -1 + 0.5 \cdot (1 - 0) = -0.5\)
-
\(\displaystyle y_2 = y_1 + 0.5 \cdot (y_1^2 - x_1) = -0.5 + 0.5 \cdot (0.25 - 0.5) = -0.625\)
Therefore, \(y(0.5) \approx -0.5\) and \(y(1) \approx -0.625\text{.}\)
8.4.2.
Solution.
📝 120. Alternative Derivation Approach.
The solution below uses a forward-difference approximation to derive Euler’s formula. This is an alternative to the geometric "slope and step size" approach used in the main text. Both methods arrive at the same formula, but this algebraic approach provides another perspective on why Euler’s method works.
Euler’s method works by replacing \(y'(x)\) in (8.5) with its forward-difference approximation, giving us
\begin{equation*}
\frac{1}{h}\Big(y(x+h) - y(x)\Big) - 2x y(x) = 0
\end{equation*}
and if we let \(h\) be the space between the \(x\)-values (0.5), then we have
\begin{equation*}
\frac{1}{0.5}\Big(y(x+0.5) - y(x)\Big) - 2x y(x) = 0.
\end{equation*}
To see how this is helpful, we isolate \(y(x+h)\) as follows:
\begin{align*}
\frac{1}{0.5}\Big(y(x+0.5) - y(x)\Big) =\ \amp 2x y(x) \\
y(x+0.5) - y(x) =\ \amp x y(x)
\end{align*}
and the final step gives the approximation formula:
\begin{equation}
y(x+0.5) = y(x) + x y(x).\tag{8.7}
\end{equation}
Note that we already know \(y(0) = 2\) from (8.6), so to find the rest of the approximation points, we plug \(x = 0, 0.5, 1, 1.5\) into (8.7), as shown below.
| \(x\) | Approximation Formula (8.7) | Approximation |
| \(0\) |
\(y(0+0.5) = y(0) + (0)y(0)\) \(\phantom{y(0+0.5)} = 2 + (0)(2) \) \(\phantom{y(0+0.5)} = 2 \) |
\(y(0.5) = 2\) |
| \(0.5\) |
\(y(0.5+0.5) = y(0.5) + (0.5)y(0.5)\) \(\phantom{y(0.5+0.5)} = 2 + (0.5)(2) \) \(\phantom{y(0.5+0.5)} = 3 \) |
\(y(1) = 3\) |
| \(1\) |
\(y(1+0.5) = y(1) + (1)y(1)\) \(\phantom{y(1+0.5)} = 3 + (1)(3) \) \(\phantom{y(1+0.5)} = 6 \) |
\(y(1.5) = 6\) |
| \(1.5\) |
\(y(1.5+0.5) = y(1.5) + (1.5)y(1.5)\) \(\phantom{y(1.5+0.5)} = 6 + (1.5)(6) \) \(\phantom{y(1.5+0.5)} = 15 \) |
\(y(2) = 15\) |
8.4.6. Conceptual Understanding.
Solution.
-
Euler’s method approximates the solution curve by following straight-line segments in the direction of the slope. When the step size is large, these straight segments can deviate significantly from the curved solution, especially if the curvature is high. Smaller steps mean the straight-line approximation stays closer to the actual curve before each correction, reducing the accumulated error.
-
This drifting behavior suggests error accumulation. Each step of Euler’s method introduces a small error (called local truncation error), and these errors can compound over many steps. To improve the approximation, you could: (1) use a smaller step size, (2) use a more accurate numerical method, or (3) check your calculations for arithmetic errors.
-
Yes, Euler’s method can produce exact solutions for certain differential equations. For example, if \(y' = c\) (constant), the solution is a straight line \(y = ct + y_0\text{,}\) and Euler’s method follows this line exactly. More generally, any differential equation whose solution is piecewise linear will be solved exactly by Euler’s method with appropriate step placement. However, for most differential equations with curved solutions, Euler’s method only produces approximations.
9 Homogeneous Equations (LHCC)
9.6 Exercises
✍🏻 Problems
9.6.7. Verifying Superposition.
Solution.
Differentiating \(y\text{:}\)
\begin{gather*}
y' = c_1 e^x + 2c_2 e^{2x} \\
y'' = c_1 e^x + 4c_2 e^{2x}
\end{gather*}
Substituting into (9.8):
\begin{gather*}
(c_1 e^x + 4c_2 e^{2x}) - 3(c_1 e^x + 2c_2 e^{2x}) + 2(c_1 e^x + c_2 e^{2x}) = 0 \\
c_1 e^x + 4c_2 e^{2x} - 3c_1 e^x - 6c_2 e^{2x} + 2c_1 e^x + 2c_2 e^{2x} = 0 \\
(c_1 e^x - 3c_1 e^x + 2c_1 e^x) + (4c_2 e^{2x} - 6c_2 e^{2x} + 2c_2 e^{2x}) = 0 \\
0 = 0.
\end{gather*}
Since the equation holds for all values of \(c_1\) and \(c_2\text{,}\) this confirms that their linear combination is also a solution.
9.6.8.
Solution.
To find the general solution, we need to identify all the roots from the factored characteristic equation:
-
\((r-1)^2 = 0\) gives \(r = 1\) with multiplicity 2
-
\((r+3) = 0\) gives \(r = -3\) with multiplicity 1
-
\((r^2 + 2r + 5)^2 = 0\) requires solving \(r^2 + 2r + 5 = 0\)
Using the quadratic formula on \(r^2 + 2r + 5 = 0\text{:}\)
\begin{equation*}
r = \frac{-2 \pm \sqrt{4-20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i
\end{equation*}
Since the quadratic factor has multiplicity 2, the complex roots \(r = -1 \pm 2i\) each have multiplicity 2.
The general solution combines:
-
For \(r = 1\) (multiplicity 2): \(c_1 e^{x} + c_2 x e^{x}\)
-
For \(r = -3\text{:}\) \(c_3 e^{-3x}\)
-
For \(r = -1 \pm 2i\) (multiplicity 2): \(e^{-x}[(c_4 + c_5 x)\cos(2x) + (c_6 + c_7 x)\sin(2x)]\)
Therefore, the general solution is:
\begin{equation*}
y = c_1 e^{x} + c_2 x e^{x} + c_3 e^{-3x} + e^{-x}[(c_4 + c_5 x)\cos(2x) + (c_6 + c_7 x)\sin(2x)]
\end{equation*}
9.6.9.
Solution.
To find the general solution, we need to identify all the roots from the factored characteristic equation:
-
\((r+1)^2 = 0\) gives \(r = -1\) with multiplicity 2
-
\((r-6)^3 = 0\) gives \(r = 6\) with multiplicity 3
-
\((r^2+1) = 0\) gives \(r = \pm i\)
-
\((r^2 + 4) = 0\) gives \(r = \pm 2i\)
The general solution combines:
-
For \(r = -1\) (multiplicity 2): \(c_1 e^{-x} + c_2 x e^{-x}\)
-
For \(r = 6\) (multiplicity 3): \(c_3 e^{6x} + c_4 x e^{6x} + c_5 x^2 e^{6x}\)
-
For \(r = \pm i\text{:}\) \(c_6 \cos(x) + c_7 \sin(x)\)
-
For \(r = \pm 2i\text{:}\) \(c_8 \cos(2x) + c_9 \sin(2x)\)
Therefore, the general solution is:
\begin{equation*}
y = c_1 e^{-x} + c_2 x e^{-x} + c_3 e^{6x} + c_4 x e^{6x} + c_5 x^2 e^{6x} + c_6 \cos(x) + c_7 \sin(x) + c_8 \cos(2x) + c_9 \sin(2x)
\end{equation*}
9.6.10.
Solution.
To find the general solution, we need to identify all the roots from the factored characteristic equation:
-
\((r-1)^2 = 0\) gives \(r = 1\) with multiplicity 2
-
\((r+3) = 0\) gives \(r = -3\) with multiplicity 1
-
\((r^2 + 2r + 5)^2 = 0\) requires solving \(r^2 + 2r + 5 = 0\)
Using the quadratic formula on \(r^2 + 2r + 5 = 0\text{:}\)
\begin{equation*}
r = \frac{-2 \pm \sqrt{4-20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i
\end{equation*}
Since the quadratic factor has multiplicity 2, the complex roots \(r = -1 \pm 2i\) each have multiplicity 2.
The general solution combines:
-
For \(r = 1\) (multiplicity 2): \(c_1 e^{x} + c_2 x e^{x}\)
-
For \(r = -3\text{:}\) \(c_3 e^{-3x}\)
-
For \(r = -1 \pm 2i\) (multiplicity 2): \(e^{-x}[(c_4 + c_5 x)\cos(2x) + (c_6 + c_7 x)\sin(2x)]\)
Therefore, the general solution is:
\begin{equation*}
y = c_1 e^{x} + c_2 x e^{x} + c_3 e^{-3x} + e^{-x}[(c_4 + c_5 x)\cos(2x) + (c_6 + c_7 x)\sin(2x)]
\end{equation*}
9.6.11.
Solution.
To find the general solution, we need to identify all the roots from the factored characteristic equation:
-
\((r+1)^2 = 0\) gives \(r = -1\) with multiplicity 2
-
\((r-6)^3 = 0\) gives \(r = 6\) with multiplicity 3
-
\((r^2+1) = 0\) gives \(r = \pm i\)
-
\((r^2 + 4) = 0\) gives \(r = \pm 2i\)
The general solution combines:
-
For \(r = -1\) (multiplicity 2): \(c_1 e^{-x} + c_2 x e^{-x}\)
-
For \(r = 6\) (multiplicity 3): \(c_3 e^{6x} + c_4 x e^{6x} + c_5 x^2 e^{6x}\)
-
For \(r = \pm i\text{:}\) \(c_6 \cos(x) + c_7 \sin(x)\)
-
For \(r = \pm 2i\text{:}\) \(c_8 \cos(2x) + c_9 \sin(2x)\)
Therefore, the general solution is:
\begin{equation*}
y = c_1 e^{-x} + c_2 x e^{-x} + c_3 e^{6x} + c_4 x e^{6x} + c_5 x^2 e^{6x} + c_6 \cos(x) + c_7 \sin(x) + c_8 \cos(2x) + c_9 \sin(2x)
\end{equation*}
9.6.12.
9.6.13.
9.6.14.
Solution.
So the roots are \(r = 0\text{,}\) \(\sqrt{3}\text{,}\) and \(-\sqrt{3}\text{.}\) Therefore:
\begin{equation*}
M(t) = c_1 + c_2 e^{\sqrt{3}t} + c_3 e^{-\sqrt{3}t}
\end{equation*}
9.6.15.
Solution.
So the roots are \(r = 0\) and \(r = 9\text{.}\) Since these are distinct real roots, the general solution is:
\begin{equation*}
y = c_1 + c_2 e^{9x}
\end{equation*}
9.6.16.
9.6.17.
Solution.
Solving for \(r\text{,}\) we get \(r = \dfrac{1 \pm \sqrt{45}}{2}\text{.}\) Therefore, the general solution is:
\begin{equation*}
y = c_1 e^{\dfrac{1 + \sqrt{45}}{2}x} + c_2 e^{\dfrac{1 - \sqrt{45}}{2}x}
\end{equation*}
9.6.18.
9.6.19.
Solution.
So we have a repeated root \(r = 1\text{.}\) Therefore, the general solution is:
\begin{equation*}
m = c_1 e^{t} + c_2 t e^{t}
\end{equation*}
9.6.20.
9.6.21.
9.6.22.
9.6.23.
9.6.24.
Solution.
Solving for \(r\text{:}\)
\begin{equation*}
r^2 = -49 \quad \Rightarrow \quad r = \pm 7i
\end{equation*}
Since we have complex roots \(r = 0 \pm 7i\) (that is, \(\alpha = 0\) and \(\beta = 7\)), the general solution is:
\begin{equation*}
w = c_1 \cos(7x) + c_2 \sin(7x)
\end{equation*}
9.6.25.
9.6.26.
9.6.27.
Solution.
Therefore, the roots are \(r = 0\) and \(r = -\dfrac{4}{3}\text{.}\) The general solution is:
\begin{equation*}
y = c_1 + c_2 e^{-\dfrac{4}{3}x}
\end{equation*}
9.6.28.
9.6.29.
9.6.30.
Solution.
The characteristic equation is:
\begin{gather*}
r^4 - 5r^2 + 4 = 0 \text{.}
\end{gather*}
Let \(u = r^2\text{,}\) then we can rewrite the equation as:
\begin{gather*}
u^2 - 5u + 4 = 0 \text{.}
\end{gather*}
Solving for \(u\text{,}\) we get:
\begin{gather*}
u = 1, \quad u = 4 \text{.}
\end{gather*}
Thus, \(r^2 = 1\) gives \(r = \pm 1\text{,}\) and \(r^2 = 4\) gives \(r = \pm 2\text{.}\) The general solution is:
\begin{equation*}
y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x}\text{.}
\end{equation*}
9.6.31.
Solution.
We already have the general solution \(\ds y = c_1e^{3t} + c_2e^{-3t} \) . In order to use the initial conditions, we will eventually need the first derivative, so let’s find that now. \(\ds y' = 3c_1e^{3t} -3 c_2e^{-3t} \) Now we can see what comes of the first initial condition \(\ds y(0) = 4. \)
\begin{align*}
y(0) \amp = 4 \\
c_1e^{3\cdot 0} + c_2e^{-3\cdot 0} \amp = 4 \\
c_1 + c_2 \amp = 4
\end{align*}
Now we can use the other initial condition \(\ds y'(0) = -6. \)
\begin{align*}
y'(0) \amp = -6 \\
3c_1e^{3\cdot 0} -3 c_2e^{-3\cdot 0} \amp = -6 \\
3c_1 - 3c_2 \amp = -6
\end{align*}
Notice that the resulting equations constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) an \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for\(\ds c_2 \) in equation (\ref{eq7}) and then substitute into equation (\ref{eq8}) as follows.
\begin{align*}
c_2 \amp = 4 - c_1 \\
3c_1 - 3(4 - c_1) \amp = -6 \\
3c_1 - 12 + 3c_1 \amp = -6 \\
6c_1 - 12 \amp = -6 \\
6c_1 \amp = 6 \\
c_1 \amp = 1 \\
c_2 \amp = 4 - 1 \\
\amp = 3
\end{align*}
Hence, we have the solution \(\ds y = e^{3t} + 3e^{-3t}. \) Note: it’s not clear whether the independent variable i \(\ds x \) o \(\ds t \) , so you could replace th \(\ds t \) ’s wit \(\ds x \) ’s.
9.6.32.
Solution.
We begin by verifying the following: the DE is linear,
the DE is homogeneous, and
the DE has constant coefficients. Since all of the conditions are true, we can safely proceed by writing the characteristic equation and then solving it (either by factoring or using the quadratic equation).
the DE is homogeneous, and
the DE has constant coefficients. Since all of the conditions are true, we can safely proceed by writing the characteristic equation and then solving it (either by factoring or using the quadratic equation).
\begin{align*}
r^2 - 4r + 4 \amp = 0 \\
(r-2)(r-2) \amp = 0 \\
r \amp = 2 \mbox{ (double root)}
\end{align*}
Since \(\ds r=2 \) is a repeated real root, the general solution is
\begin{align*}
z \amp = c_1e^{2x} + c_2xe^{2x} \\
\amp = (c_1 + c_2x)e^{2x}
\end{align*}
In order to use the initial conditions, we will eventually need the first derivative, so let’s find that now. Note that we will use the product rule to take the derivative.
\begin{align*}
z' \amp = (c_1+ c_2x)\cdot 2e^{2x} + c_2 \cdot e^{2x} \\
\amp = (2c_1+ 2c_2x)e^{2x} + c_2e^{2x} \\
\amp = (2c_1 + 2c_2x + c_2)e^{2x}
\end{align*}
Now we can see what comes of the first initial condition \(\ds z(1) = 1. \)
\begin{align*}
z(1) \amp = 1 \\
(c_1 + c_2\cdot 1)e^{2\cdot 1} \amp = 1 \\
(c_1 + c_2)e^{2} \amp = 1 \\
c_1 + c_2 \amp = \dfrac{1}{e^2} \\
c_1 + c_2 \amp = e^{-2}
\end{align*}
Now we can use the other initial condition \(\ds z'(1) = 1. \)
\begin{align*}
z'(1) \amp = 1 \\
(2c_1 + 2c_2\cdot 1 + c_2)e^{2\cdot 1} \amp = 1 \\
(2c_1 + 3c_2)e^{2} \amp = 1 \\
2c_1 + 3c_2 \amp = \dfrac{1}{e^2} \\
2c_1 + 3c_2 \amp = e^{-2}
\end{align*}
Notice that the resulting equations, (\ref{eq13}) and (\ref{eq14}) constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) and \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for \(\ds c_2 \) in equation (\ref{eq13}) and then substitute into equation (\ref{eq14}) as follows.
\begin{align*}
c_2 \amp = e^{-2} - c_1 \\
2c_1 +3(e^{-2} - c_1) \amp = e^{-2} \\
2c_1 +3e^{-2} - 3c_1 \amp = e^{-2} \\
-c_1 + 3e^{-2} \amp = e^{-2} \\
-c_1 \amp = e^{-2} - 3e^{-2} \\
-c_1 \amp = - 2e^{-2} \\
c_1 \amp = 2e^{-2} \\
c_2 \amp = e^{-2} - 2e^{-2} \\
\amp = -e^{-2}
\end{align*}
Hence, we have the solution
\begin{align*}
z \amp = (2e^{-2} - e^{-2}x)e^{2x}, \mbox{ or} \\
\amp = (2 - x)e^{-2}e^{2x}, \mbox{ or} \\
\amp = (2 - x)e^{-2 + 2x}, \mbox{ or} \\
\amp = (2 - x)e^{2x - 2}.
\end{align*}
9.6.33.
Solution.
We already have the general solution \(\ds \theta = c_1e^{2t} + c_2e^{-2t} \) . In order to use the initial conditions, we will eventually need the first derivative, so let’s find that now. \(\ds \theta' = 2c_1e^{2t} -2 c_2e^{-2t} \) Now we can see what comes of the first initial condition \(\ds \theta(0) = 12. \)
\begin{align*}
\theta(0) \amp = 12 \\
c_1e^{2\cdot 0} + c_2e^{-2\cdot 0} \amp = 12 \\
c_1 + c_2 \amp = 12
\end{align*}
Now we can use the other initial condition \(\ds \theta'(0) = -2. \)
\begin{align*}
\theta'(0) \amp = -2 \\
2c_1e^{2\cdot 0} -2 c_2e^{-2\cdot 0} \amp = -2 \\
2c_1 - 2c_2 \amp = -2
\end{align*}
Notice that the resulting equations, (\ref{eq17}) and (\ref{eq18}) constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) and \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for \(\ds c_2 \) in equation (\ref{eq17}) and then substitute into equation (\ref{eq18}) as follows.
\begin{align*}
c_2 \amp = 12 - c_1 \\
2c_1 - 2(12 - c_1) \amp = -2 \\
2c_1 - 24 + 2c_1 \amp = -2 \\
4c_1 - 24 \amp = -2 \\
4c_1 \amp = 22 \\
c_1 \amp = \dfrac{22}{4} \\
c_1 \amp = \dfrac{11}{2} \\
c_2 \amp = 12 - \dfrac{11}{2} \\
\amp = \dfrac{13}{2}
\end{align*}
Hence, we have the solution \(\ds \theta = \dfrac{11}{2}e^{2t} + \dfrac{13}{2}e^{-2t}. \) Note: it’s not clear whether the independent variable i \(\ds x \) o \(\ds t \) , so you could replace th \(\ds t \) ’s wit \(\ds x \) ’s.
9.6.34. The General Solution to First-Order LHCC Equations.
Solution.
The characteristic equation this equation is
\begin{equation*}
a_1\ r + a_0 = 0 \quad \Rightarrow \quad r = -a_0/a_1\text{.}
\end{equation*}
So, the only fundamental solution to this equation is:\(\ y = e^{-(\sfrac{a_0}{a_1}) x}\text{.}\)
Therefore, the general solution must be
\begin{equation*}
y = c_1 e^{-(\sfrac{a_0}{a_1}) x}
\end{equation*}
where \(c_1\) is an arbitrary constant.
9.6.36.
Solution.
We begin by verifying the following: the DE is linear,
the DE is homogeneous, and
the DE has constant coefficients. Since all of the conditions are true, we can safely proceed to find the general solution. We write down and then solve the characteristic equation, as follows.
the DE is homogeneous, and
the DE has constant coefficients. Since all of the conditions are true, we can safely proceed to find the general solution. We write down and then solve the characteristic equation, as follows.
\begin{align*}
r^2 - 1 \amp = 0 \\
r^2 \amp = 1 \\
r \amp = \pm \sqrt{1} \\
r \amp = \pm 1
\end{align*}
Since \(\ds r = 1 \) an \(\ds r = -1 \) are distinct, real roots, the general solution is given by \(\ds y = c_1e^{t} + c_2e^{-t}. \) Now we can see what comes of the first boundary condition \(\ds y(0) = 1. \)
\begin{align*}
y(0) \amp = 1 \\
c_1e^{0} + c_2e^{-0} \amp = 1 \\
c_1 + c_2 \amp = 1
\end{align*}
Now we can use the other boundary condition \(\ds y(1) = 2e- \dfrac{1}{e}. \)
\begin{align*}
y(1) \amp = 2e- \dfrac{1}{e} \\
c_1e^{1} + c_2e^{-1} \amp = 2e- \dfrac{1}{e} \\
c_1\cdot e + c_2\cdot \dfrac{1}{e} \amp = 2e- \dfrac{1}{e}
\end{align*}
Notice that the resulting equations, (\ref{eq11}) and (\ref{eq12}) constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) an \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for\(\ds c_2 \) in equation (\ref{eq11}) and then substitute into equation (\ref{eq12}) as follows.
\begin{align*}
c_2 \amp = 1 - c_1 \\
c_1\cdot e + (1 - c_1)\cdot \dfrac{1}{e} \amp = 2e- \dfrac{1}{e} \\
c_1\cdot e + \dfrac{1}{e} - c_1\cdot \dfrac{1}{e} \amp = 2e- \dfrac{1}{e} \\
c_1\cdot e - c_1\cdot \dfrac{1}{e} \amp = 2e - \dfrac{2}{e} \\
c_1\left(e - \dfrac{1}{e} \right) \amp = 2e - \dfrac{2}{e} \\
c_1 \amp = \dfrac{2e - \dfrac{2}{e}}{e - \dfrac{1}{e}} \\
\amp = \dfrac{2e - \dfrac{2}{e}}{e - \dfrac{1}{e}} \cdot \dfrac{e}{e} \\
\amp = \dfrac{2e^2 - 2}{e^2 - 1} \\
\amp = \dfrac{2(e^2 - 1)}{e^2 - 1} \\
\amp = 2 \\
c_2 \amp = 1 - 2 \\
\amp = -1
\end{align*}
Hence, we have the solution \(\ds y = 2e^{t} - e^{-t}. \)
9.6.37.
Solution.
Consider a second-order LHCC equation with a repeated root \(r\) in its characteristic equation.
From the characteristic equation, we know that \(y_1 = e^{rx}\) is one solution to the differential equation. However, for a second-order equation, we need two linearly independent solutions to form the general solution.
If we try to use \(y = c_1 e^{rx} + c_2 e^{rx}\text{,}\) we can factor:
\begin{equation*}
y = (c_1 + c_2) e^{rx} = C e^{rx}
\end{equation*}
where \(C = c_1 + c_2\text{.}\) This gives us only one independent solution, not two.
To find a second, independent solution, we multiply the exponential by \(x\text{,}\) giving \(y_2 = xe^{rx}\text{.}\) We can verify this works by substituting into the differential equation. For example, if the DE is \(y'' - 2ry' + r^2y = 0\text{:}\)
\begin{align*}
y_2 \amp = xe^{rx}\\
y_2' \amp = e^{rx} + rxe^{rx}\\
y_2'' \amp = re^{rx} + re^{rx} + r^2xe^{rx} = 2re^{rx} + r^2xe^{rx}
\end{align*}
Substituting:
\begin{align*}
\amp (2re^{rx} + r^2xe^{rx}) - 2r(e^{rx} + rxe^{rx}) + r^2(xe^{rx})\\
\amp = 2re^{rx} + r^2xe^{rx} - 2re^{rx} - 2r^2xe^{rx} + r^2xe^{rx}\\
\amp = 0
\end{align*}
Thus, \(y_2 = xe^{rx}\) is indeed a solution. Moreover, \(y_1 = e^{rx}\) and \(y_2 = xe^{rx}\) are linearly independent because one is not a constant multiple of the other. Therefore, the general solution is:
\begin{equation*}
y = c_1 e^{rx} + c_2 xe^{rx}
\end{equation*}
10 Nonhomogeneous Equations (LNCC)
10.4 Exercises
💡 Conceptual Quiz
10.4.1.
10.4.1.n
Solution.
A differential equation is termed "homogeneous" if all the terms of the equation are either derivatives of the unknown function or multiples of the unknown function itself, and the equation can be set equal to zero. In other words, there are no terms in the equation that are solely functions of the independent variable or constants.
For example, the equation \(y'' + 2y' + y = 0 \) is homogeneous because all its terms involve \(y \) or its derivatives, and the equation is set equal to zero.
A differential equation is termed "non-homogeneous" if it contains terms that are solely functions of the independent variable or constants that are not multiples of the unknown function or its derivatives.
For example, the equation \(y'' + 2y' + y = e^x \) is non-homogeneous because of the term \(e^x \) on the right side, which is not a derivative or multiple of \(y \text{.}\)
The distinction between homogeneous and non-homogeneous differential equations is crucial, as the methods used to solve them can differ.
🏋️♂️ Practice Drills
10.4.11.
Solution.
First, solve the homogeneous equation \(3z'' - 4z' - 12z = 0\) to find \(y_c\text{.}\) The characteristic equation is:
\begin{align*}
3r^2 - 4r - 12 \amp = 0 \\
r \amp = \frac{2 \pm 2\sqrt{10}}{3}
\end{align*}
Since the roots are distinct and real, \(z_c = c_1 e^{\frac{2 + 2\sqrt{10}}{3}\theta} + c_2 e^{\frac{2 - 2\sqrt{10}}{3}\theta}\text{.}\)
The forcing term is \(17 + 2\cos(2\theta)\text{,}\) a constant plus a trig function. Our initial guess is:
\begin{equation*}
z_p = A + B\cos(2\theta) + C\sin(2\theta)
\end{equation*}
No duplicates with \(z_c\) occur, so this guess stands.
10.4.12.
Solution.
Solve the homogeneous equation \(y'' - 3y' - 17y = 0\text{.}\)
\begin{align*}
r^2 - 3r - 17 \amp = 0 \\
r \amp = \frac{3 \pm \sqrt{77}}{2}
\end{align*}
Distinct real roots give \(y_c = c_1 e^{\frac{3 + \sqrt{77}}{2}x} + c_2 e^{\frac{3 - \sqrt{77}}{2}x}\text{.}\)
The forcing term is a 2nd-degree polynomial plus a cosine. Our guess is:
\begin{equation*}
y_p = (Ax^2 + Bx + C) + (D\cos x + E\sin x)
\end{equation*}
None of these terms repeat \(y_c\text{.}\)
10.4.13.
Solution.
Solve \(y'' + y' - 12y = 0\text{.}\)
\begin{align*}
r^2 + r - 12 \amp = 0 \\
(r+4)(r-3) \amp = 0 \\
r \amp = -4, 3
\end{align*}
So, \(y_c = c_1 e^{-4t} + c_2 e^{3t}\text{.}\)
The forcing term has \(4e^{3t}\) and \(\cos(2t)\text{.}\) Our first guess is:
\begin{equation*}
y_p = Ae^{3t} + B\cos(2t) + C\sin(2t)
\end{equation*}
Since \(e^{3t}\) appears in \(y_c\text{,}\) we multiply that term by \(t\text{:}\)
\begin{equation*}
y_p = Ate^{3t} + B\cos(2t) + C\sin(2t)
\end{equation*}
10.4.14.
Solution.
Solve \(y'' - 4y = 0\text{.}\)
\begin{align*}
r^2 - 4 \amp = 0 \\
r \amp = \pm 2
\end{align*}
Distinct real roots give \(y_c = c_1 e^{2t} + c_2 e^{-2t}\text{.}\)
The forcing term is \(2t\) (a first-degree polynomial) and \(e^{2t}\sin(3t)\text{.}\) Guess:
\begin{equation*}
y_p = (At + B) + e^{2t}\big(C\cos(3t) + D\sin(3t)\big)
\end{equation*}
No repeats with \(y_c\text{.}\)
10.4.15.
Solution.
Solve \(w'' - 4w' + 13w = 0\text{.}\)
\begin{align*}
r^2 - 4r + 13 \amp = 0 \\
r \amp = 2 \pm 3i
\end{align*}
Complex roots give \(w_c = c_1 e^{2x}\cos(3x) + c_2 e^{2x}\sin(3x)\text{.}\)
The forcing term has \(e^{3x}\) and \(x^2e^{-x}\text{.}\) Guess:
\begin{equation*}
w_p = Ae^{3x} + (Bx^2 + Cx + D)e^{-x}
\end{equation*}
None of these terms appear in \(w_c\text{.}\)
10.4.16.
Solution.
Solve \(y'' - 2y' + y = 0\text{.}\)
\begin{align*}
r^2 - 2r + 1 \amp = 0 \\
r \amp = 1 \text{ (double root)}
\end{align*}
Double root gives \(y_c = c_1 e^{x} + c_2 x e^{x}\text{.}\)
Since \(e^x\) appears in \(y_c\text{,}\) modify \(De^x\) → \(Dxe^x\text{.}\) But \(xe^x\) is also in \(y_c\text{,}\) so we multiply again: \(D x^2 e^x\text{.}\)
10.4.17.
Solution.
Solve \(z'' - 4z' + 5z = 0\text{.}\) The characteristic equation is:
\begin{align*}
r^2 - 4r + 5 \amp = 0 \\
r \amp = 2 \pm i
\end{align*}
Complex roots give \(z_c = c_1 e^{2t}\cos t + c_2 e^{2t}\sin t\text{.}\)
Forcing term: \(te^{2t} + 2e^{2t}\sin t\text{.}\) First guess:
\begin{equation*}
z_p = (At + B)e^{2t} + e^{2t}(C\cos t + D\sin t)
\end{equation*}
The \(e^{2t}\cos t\) and \(e^{2t}\sin t\) terms duplicate \(y_c\text{,}\) so we multiply that piece by \(t\text{:}\)
\begin{equation*}
z_p = (At + B)e^{2t} + e^{2t}(Ct\cos t + Dt\sin t)
\end{equation*}
10.4.19.
Solution.
First solve the homogeneous equation \(y'' - 4y' - 12y = 0\text{:}\)
\begin{align*}
r^2 - 4r - 12 \amp = 0 \\
(r - 6)(r + 2) \amp = 0 \\
r \amp = -2, 6
\end{align*}
Thus, \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
Substitute \(y_p\) into the DE:
\begin{align*}
25A e^{5t} - 4(5A e^{5t}) - 12(A e^{5t}) \amp = 3 e^{5t} \\
-7A e^{5t} \amp = 3 e^{5t}
\end{align*}
Matching coefficients gives \(A = -\frac{3}{7}\text{.}\)
10.4.20.
Solution.
The homogeneous equation is the same as above: \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
Differentiate and substitute into the DE, collecting coefficients of \(\sin(2t)\) and \(\cos(2t)\text{.}\) Solve for \(A\) and \(B\text{.}\) The system yields:
\begin{equation*}
A = \frac{1}{40}, \quad B = -\frac{1}{20}
\end{equation*}
10.4.21.
Solution.
Again, \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
The forcing term is a cubic polynomial. Guess:
\begin{equation*}
y_p = A t^3 + B t^2 + C t + D\text{.}
\end{equation*}
Differentiate, substitute into the DE, and match coefficients. Solving yields:
\begin{equation*}
A = -\frac{1}{6}, \; B = \frac{1}{6}, \; C = -\frac{1}{9}, \; D = -\frac{5}{27}\text{.}
\end{equation*}
So:
\begin{equation*}
y_p = -\frac{1}{6} t^3 + \frac{1}{6} t^2 - \frac{1}{9} t - \frac{5}{27}
\end{equation*}
10.4.22.
Solution.
The homogeneous solution is \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
A naive guess would be \(y_p = A e^{6t}\text{,}\) but \(e^{6t}\) already appears in \(y_c\text{.}\)
Modify guess to \(y_p = At e^{6t}\text{.}\)
✍🏻 Problems
10.4.1.
Solution.
Step 1. Solve the homogeneous equation \(2x' + x = 0\text{:}\)
\begin{align*}
2r + 1 \amp = 0\\
r \amp = -\frac{1}{2}
\end{align*}
So \(x_h = c_1 e^{-\frac{1}{2} t}\text{.}\)
Step 2. The forcing function \(3t^2\) is a polynomial of degree 2. Guess \(x_p = At^2 + Bt + C\text{.}\)
Step 3. No overlap with \(x_h\text{,}\) so no modification needed.
Step 4. Substitute \(x_p\) into the DE:
\begin{align*}
x_p' = 2At + B\\
2(2At + B) + (At^2 + Bt + C) \amp = 3t^2\\
At^2 + (4A+B)t + (2B+C) \amp = 3t^2
\end{align*}
Matching coefficients: \(A = 3\text{,}\) \(B = -12\text{,}\) \(C = 24\text{.}\) So \(x_p = 3t^2 - 12t + 24\text{.}\)
Step 5. Write the general solution:
\begin{equation*}
x(t) = c_1 e^{-\frac{1}{2} t} + 3t^2 - 12t + 24
\end{equation*}
10.4.2.
Solution.
Step 1: Since the right-hand side is a polynomial of degree 1, we guess:
\begin{equation*}
y_p = Ax + B \text{.}
\end{equation*}
Step 2: Differentiate the guessed solution:
\begin{gather*}
y_p' = A, \quad y_p'' = 0
\end{gather*}
Step 3: Substitute into the original equation:
\begin{gather*}
0 - 3A + 2(Ax + B) = 5x + 1 \\
2Ax + 2B - 3A = 5x + 1
\end{gather*}
Step 4: Collect like-terms and solve for \(A \) and \(B \text{:}\)
\begin{gather*}
2A = 5 \quad \implies \quad A = \frac{5}{2} \\
2B - 3A = 1 \quad \implies \quad 2B - 3\left(\frac{5}{2}\right) = 1 \quad \implies \quad B = \frac{11}{2}
\end{gather*}
Step 5: Write the particular solution:
\begin{equation*}
y_p = \frac{5}{2}x + \frac{11}{2} \text{.}
\end{equation*}
10.4.3.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 4y' - 5y = 0\text{:}\)
\begin{align*}
r^2 - 4r - 5 \amp = 0\\
(r-5)(r+1) \amp = 0\\
r \amp = 5,\,-1
\end{align*}
So \(y_h = c_1 e^{5t} + c_2 e^{-t}\text{.}\)
Step 2. The forcing function is \(t + 2e^{-t}\text{.}\) Guess \(y_p = (At + B) + C e^{-t}\text{.}\)
Step 3. Because \(e^{-t}\) appears in \(y_h\text{,}\) multiply by \(t\text{:}\) \(y_p = (At + B) + C t e^{-t}\text{.}\)
Step 4. Substitute into the DE:
\begin{gather*}
y_p' = A + (-Ct + C) e^{-t}\\
y_p'' = (Ct - 2C) e^{-t}
\end{gather*}
Plugging in and matching coefficients gives: \(A = -\frac{1}{5}\text{,}\) \(B = \frac{4}{25}\text{,}\) \(C = -\frac{1}{3}\text{.}\)
Step 5. Write the general solution:
\begin{equation*}
y(t) = c_1 e^{5t} + c_2 e^{-t} - \frac{1}{5}t + \frac{4}{25} - \frac{1}{3} t e^{-t}
\end{equation*}
10.4.4.
Solution.
Step 1. Solve the homogeneous equation \(z'' - 6z' + 34z = 0\text{:}\)
\begin{align*}
r^2 - 6r + 34 \amp = 0\\
r = 3 \pm 5i
\end{align*}
So \(z_h = c_1 e^{3x} \cos 5x + c_2 e^{3x} \sin 5x\text{.}\)
Step 2. The forcing function is \(650 \sin(6x)\text{.}\) Guess \(z_p = A \cos 6x + B \sin 6x\text{.}\)
Step 3. No overlap with \(z_h\text{,}\) so no modification needed.
Step 4. Substitute into the DE:
\begin{gather*}
z_p' = -6A \sin 6x + 6B \cos 6x\\
z_p'' = -36A \cos 6x - 36B \sin 6x
\end{gather*}
Matching coefficients from the DE gives \(A = 18\) and \(B = -1\text{.}\)
Step 5. Write the general solution:
\begin{equation*}
z(x) = c_1 e^{3x} \cos 5x + c_2 e^{3x} \sin 5x + 18 \cos 6x - \sin 6x
\end{equation*}
10.4.5.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 4y' + 4y = 0\text{:}\)
\begin{align*}
r^2 - 4r + 4 \amp = 0\\
(r-2)^2 \amp = 0\\
r \amp = 2 \text{ (double root)}
\end{align*}
So \(y_h = c_1 e^{2x} + c_2 x e^{2x}\text{.}\)
Step 2. The forcing function is \(8\cos x + 12 e^{2x}\text{.}\) Guess \(y_p = (A \cos x + B \sin x) + C e^{2x}\text{.}\)
Step 3. Because \(e^{2x}\) appears in \(y_h\text{,}\) multiply by \(x\text{,}\) then check again. \(y_p = (A \cos x + B \sin x) + C x e^{2x}\text{.}\) Still overlap (the term \(x e^{2x}\) is also in \(y_h\)), so multiply by another \(x\text{:}\) \(y_p = (A \cos x + B \sin x) + C x^2 e^{2x}\text{.}\)
Step 4. Substitute into the DE, collect terms, and match coefficients:
\begin{gather*}
3A - 4B = 8\\
4A + 3B = 0\\
2C = 12
\end{gather*}
Solve: \(C = 6\text{,}\) \(A = \frac{8}{9}\text{,}\) \(B = -\frac{32}{27}\text{.}\) So \(y_p = \frac{8}{9}\cos x - \frac{32}{27}\sin x + 6x^2 e^{2x}\text{.}\)
Step 5. Write the general solution:
\begin{equation*}
y(x) = c_1 e^{2x} + c_2 x e^{2x} + \frac{8}{9}\cos x - \frac{32}{27}\sin x + 6x^2 e^{2x}
\end{equation*}
10.4.6.
Solution.
Step 1. Solve the homogeneous equation \(y'' - y' - 6y = 0\text{:}\)
\begin{align*}
r^2 - r - 6 \amp = 0\\
(r-3)(r+2) \amp = 0\\
r \amp = 3,\,-2
\end{align*}
So \(y_h = c_1 e^{3t} + c_2 e^{-2t}\text{.}\)
Step 2. The forcing function is \(2t + 3e^{3t} - e^{-2t}\text{.}\) Guess \(y_p = (At + B) + C e^{3t} + D e^{-2t}\text{.}\)
Step 3. Both \(e^{3t}\) and \(e^{-2t}\) appear in \(y_h\text{,}\) so multiply those terms by \(t\text{:}\) \(y_p = (At + B) + Ct e^{3t} + Dt e^{-2t}\text{.}\)
Step 4. Substitute into the DE, collect terms, and match coefficients:
\begin{gather*}
-6A = 2 \implies A = -\frac{1}{3}\\
-A - 6B = 0 \implies B = \frac{1}{18}\\
5C = 3 \implies C = \frac{3}{5}\\
-5D = -1 \implies D = \frac{1}{5}
\end{gather*}
So \(y_p = -\frac{1}{3}t + \frac{1}{18} + \frac{3}{5}t e^{3t} + \frac{1}{5} t e^{-2t}\text{.}\)
Step 5. Write the general solution:
\begin{equation*}
y(t) = c_1 e^{3t} + c_2 e^{-2t} - \frac{1}{3}t + \frac{1}{18} + \frac{3}{5}t e^{3t} + \frac{1}{5} t e^{-2t}
\end{equation*}
10.4.7.
Solution.
Step 1. Solve the homogeneous equation \(y'' + 3y' + 2y = 0\text{:}\)
\begin{align*}
r^2 + 3r + 2 \amp = 0\\
(r+1)(r+2) \amp = 0\\
r \amp = -1,\,-2
\end{align*}
So \(y_h = c_1 e^{-x} + c_2 e^{-2x}\text{.}\)
Step 3. No overlap with \(y_h\text{,}\) so no modification is needed.
Step 4. Substitute \(y_p\) into the DE:
\begin{align*}
y_p' = 2A e^{2x}, \ y_p'' = 4A e^{2x}\\
4A e^{2x} + 3(2A e^{2x}) + 2(A e^{2x}) \amp = 6 e^{2x}\\
12A e^{2x} \amp = 6 e^{2x}\\
A \amp = \frac{1}{2}
\end{align*}
So \(y_p = \frac{1}{2} e^{2x}\text{.}\)
Step 5. General solution:
\begin{equation*}
y(x) = c_1 e^{-x} + c_2 e^{-2x} + \frac{1}{2} e^{2x}
\end{equation*}
10.4.8.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 3y' + y = 0\text{:}\)
\begin{align*}
r^2 - 3r + 1 \amp = 0\\
r = \frac{3 \pm \sqrt{5}}{2}
\end{align*}
So \(y_h = c_1 e^{\frac{3+\sqrt{5}}{2} x} + c_2 e^{\frac{3-\sqrt{5}}{2} x}\text{.}\)
Step 2. Forcing function is a degree‑2 polynomial. Guess \(y_p = Ax^2 + Bx + C\text{.}\)
Step 3. No overlap with \(y_h\text{,}\) so no modification needed.
Step 4. Substitute into the DE and match coefficients:
\begin{align*}
y_p' = 2Ax + B, \ y_p'' = 2A\\
2A - 3(2Ax + B) + (Ax^2 + Bx + C) \amp = 2x^2 + 3x\\
Ax^2 + (-6A + B)x + (2A - 3B + C) \amp = 2x^2 + 3x
\end{align*}
Compare coefficients: \(A = 2\text{,}\) \(-6A + B = 3 \Rightarrow B = 15\text{,}\) \(2A - 3B + C = 0 \Rightarrow C = 41\text{.}\) So \(y_p = 2x^2 + 15x + 41\text{.}\)
Step 5. General solution:
\begin{equation*}
y(x) = c_1 e^{\frac{3+\sqrt{5}}{2} x} + c_2 e^{\frac{3-\sqrt{5}}{2} x} + 2x^2 + 15x + 41
\end{equation*}
10.4.9.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 4y' - 12y = 0\text{:}\)
\begin{align*}
r^2 - 4r - 12 \amp = 0\\
r = 6,\ -2
\end{align*}
So \(y_h = c_1 e^{6t} + c_2 e^{-2t}\text{.}\)
Step 4. Substitute into the DE:
\begin{gather*}
y_p' = A e^{4t} + (4At + 4B) e^{4t}\\
y_p'' = 4A e^{4t} + (4A + 16B + 16At) e^{4t}
\end{gather*}
Collect like terms, match coefficients, and solve to find: \(A = \frac{1}{4}\text{,}\) \(B = 0\text{.}\) So \(y_p = \frac{1}{4} t e^{4t}\text{.}\)
Step 5. General solution:
\begin{equation*}
y(t) = c_1 e^{6t} + c_2 e^{-2t} + \frac{1}{4} t e^{4t}
\end{equation*}
10.4.10.
Solution.
Step 1. Solve the homogeneous equation \(y'' + y = 0\text{:}\)
\begin{align*}
r^2 + 1 \amp = 0\\
r = \pm i
\end{align*}
So \(y_h = c_1 \cos x + c_2 \sin x\text{.}\)
Step 3. Both \(\cos x\) and \(\sin x\) appear in \(y_h\text{,}\) so multiply guess by \(x\text{:}\) \(y_p = x(A \cos x + B \sin x)\text{.}\)
Step 4. Substitute into the DE and match coefficients to find: \(A = 0\text{,}\) \(B = \frac{3}{2}\text{.}\)
Step 5. General solution:
\begin{equation*}
y(x) = c_1 \cos x + c_2 \sin x + \frac{3}{2} x \sin x
\end{equation*}
10.4.11.
Solution.
Step 1. Solve the homogeneous equation \(y'' + 3y' - 28y = 0\text{:}\)
\begin{align*}
r^2 + 3r - 28 \amp = 0\\
(r+7)(r-4) \amp = 0\\
r \amp = -7,\ 4
\end{align*}
So \(y_h = c_1 e^{-7t} + c_2 e^{4t}\text{.}\)
Step 2. Forcing function is \(7t + e^{4t} - 1\text{.}\) Guess \(y_p = (At + B) + C e^{4t}\text{.}\)
Step 3. Because \(e^{4t}\) appears in \(y_h\text{,}\) multiply that term by \(t\text{:}\) \(y_p = (At + B) + C t e^{4t}\text{.}\)
Step 4. Substitute into the DE, collect terms, and solve for coefficients: \(A = -\frac{1}{4}\text{,}\) \(B = \frac{1}{112}\text{,}\) \(C = \frac{1}{11}\text{.}\)
Step 5. General solution:
\begin{equation*}
y(t) = c_1 e^{-7t} + c_2 e^{4t} - \frac{1}{4}t + \frac{1}{112} + \frac{1}{11} t e^{4t}
\end{equation*}
10.4.12.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 100y = 0\text{:}\)
\begin{align*}
r^2 - 100 \amp = 0\\
r = \pm 10
\end{align*}
So \(y_h = c_1 e^{10t} + c_2 e^{-10t}\text{.}\)
Step 2. The forcing function is a combination:
-
\(9 t^2 e^{10t}\) (poly × exponential)
-
\(\cos t\) (trig)
-
\(- t \sin t\) (poly × trig)
We guess each piece separately and combine:
\begin{gather*}
y_{p1} = (A t^3 + B t^2 + C t) e^{10t} \text{ (because e^{10t} is in y_h, multiply by t)}\\
y_{p2} = D \cos t + E \sin t\\
y_{p3} = (Ft + G)\sin t + (Ht + I)\cos t
\end{gather*}
Then form \(y_p = y_{p1} + y_{p2} + y_{p3}\text{.}\)
Step 4. Substitute into the DE. Match coefficients for each independent function:
-
From \(9 t^2 e^{10t}\text{,}\) solve for \(A, B, C\text{.}\)
-
From \(\cos t\) and \(\sin t\text{,}\) solve for \(D, E, F, G, H, I\text{.}\)
(Algebra omitted for brevity here, but systematically done.)
Step 5. General solution:
\begin{equation*}
y(t) = c_1 e^{10t} + c_2 e^{-10t} + (A t^3 + B t^2 + C t) e^{10t} + D \cos t + E \sin t + (Ft + G)\sin t + (Ht + I)\cos t
\end{equation*}
10.4.13.
Solution.
Step 1. Homogeneous equation: \(4y'' + y = 0\text{.}\)
\begin{align*}
4r^2 + 1 \amp = 0\\
r = \pm \frac{i}{2}
\end{align*}
So \(y_h = c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2}\text{.}\)
Step 2. Forcing has two pieces:
-
\(e^{-2t} \sin \left( \frac{t}{2} \right)\) → guess \((A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t}\)
-
\(6 t \cos \left( \frac{t}{2} \right)\) → guess \((C t^2 + D t) \cos \frac{t}{2} + (E t^2 + F t) \sin \frac{t}{2}\)
Step 3. The term \(6 t \cos \frac{t}{2}\) overlaps with \(y_h\text{,}\) so multiply that guess by \(t\text{.}\)
Step 4. Substitute all into the DE, match coefficients for each independent function, and solve for \(A \ldots F\text{.}\)
Step 5. General solution:
\begin{equation*}
y(t) = c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2} + (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t} + (C t^2 + D t) \cos \frac{t}{2} + (E t^2 + F t) \sin \frac{t}{2}
\end{equation*}
10.4.14.
Solution.
Step 1. Homogeneous equation: \(4y'' + 16y' + 17y = 0\text{.}\)
\begin{align*}
4r^2 + 16r + 17 \amp = 0\\
r = -2 \pm \frac{i}{2}
\end{align*}
So \(y_h = e^{-2t}\big(c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2}\big)\text{.}\)
Step 2. Forcing has two pieces, same type as previous problem. Guess terms similar to uc-gen-12 but multiply by \(t\) if overlap occurs:
\begin{gather*}
y_{p1} = t (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t}\\
y_{p2} = (C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2}
\end{gather*}
Step 4. Substitute and match coefficients for all six unknowns.
Step 5. General solution:
\begin{equation*}
y(t) = e^{-2t}\big(c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2}\big) + t (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t} + (C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2}
\end{equation*}
10.4.15.
Solution.
Step 1. Solve homogeneous equation \(y'' + 8y' + 16y = 0\text{:}\)
\begin{align*}
r^2 + 8r + 16 \amp = 0\\
(r+4)^2 \amp = 0\\
r \amp = -4 \text{ (double root)}
\end{align*}
So \(y_h = c_1 e^{-4t} + c_2 t e^{-4t}\text{.}\)
Step 2. Forcing function is \(e^{-4t} + (t^2 + 5) e^{-4t}\text{.}\) Guess \(y_p = (A t^4 + B t^3 + C t^2) e^{-4t}\) because of double overlap (every term has \(e^{-4t}\)).
Step 4. Substitute, match coefficients and solve for \(A,B,C\text{.}\)
Step 5. General solution:
\begin{equation*}
y(t) = c_1 e^{-4t} + c_2 t e^{-4t} + (A t^4 + B t^3 + C t^2) e^{-4t}
\end{equation*}
10.4.16.
Solution.
Step 1. Homogeneous: \(y_h = C e^{3x}\text{.}\)
Step 2. Forcing is constant 6 → guess \(y_p = A\text{.}\)
Step 3. No overlap.
Step 4. Substitute: \(-3A = 6 → A = -2\text{.}\)
Step 5. General: \(y = C e^{3x} - 2\text{.}\)
Step 6. Use \(y(0)=5\text{:}\) \(C = 7\text{.}\)
\begin{equation*}
y = 7e^{3x} - 2\text{.}
\end{equation*}
10.4.17.
Solution.
Step 1. Homog: \(2x' + x=0 → x_h = C e^{-t/2}\text{.}\)
Step 2. Guess \(x_p = At^2 + Bt + C_0\) (poly deg 2).
Step 3. No overlap.
Step 5. General: \(x = C e^{-t/2} + 3t^2 -12t + 24\text{.}\)
Step 6. Use \(x(0)=15\text{:}\) \(C=-9\text{.}\)
\begin{equation*}
x(t) = -9e^{-t/2} + 3t^2 -12t + 24\text{.}
\end{equation*}
10.4.18.
Solution.
Step 2. Guess \(z_p = A e^{-t}\text{.}\)
Step 3. No overlap.
Step 4. Substitute: \(3A e^{-t} = e^{-t} → A=\frac{1}{3}\text{.}\)
Step 5. General: \(z = C e^{-4t} + \frac{1}{3} e^{-t}\text{.}\)
Step 6. Apply \(z(0)=\frac{4}{3}\text{:}\) \(C=1\text{.}\)
\begin{equation*}
z(t) = e^{-4t} + \frac{1}{3} e^{-t}\text{.}
\end{equation*}
10.4.19.
Solution.
Step 1. Homog: \(2z'' + z=0 → r = \pm \frac{i}{\sqrt{2}}\text{.}\) \(z_h = c_1 \cos \frac{t}{\sqrt{2}} + c_2 \sin \frac{t}{\sqrt{2}}\text{.}\)
Step 2. Guess \(z_p = A e^{2t}\text{.}\)
Step 3. No overlap.
Step 4. Substitute: \(9A e^{2t} = 9 e^{2t} → A=1\text{.}\)
Step 5. General:
\begin{equation*}
z = c_1 \cos \frac{t}{\sqrt{2}} + c_2 \sin \frac{t}{\sqrt{2}} + e^{2t}\text{.}
\end{equation*}
Step 6. Apply ICs:
\begin{gather*}
3 = c_1 + 1 → c_1 = 2\\
-1 = \frac{1}{\sqrt{2}}c_2 + 2 → c_2 = -3\sqrt{2}
\end{gather*}
So \(z(t) = 2\cos \frac{t}{\sqrt{2}} - 3\sqrt{2} \sin \frac{t}{\sqrt{2}} + e^{2t}\text{.}\)
10.4.20.
Solution.
Step 1. Homog: \(y'' - 4y' - 12y=0 → r=6,-2 → y_h=c_1 e^{6t} + c_2 e^{-2t}\text{.}\)
Step 2. Guess \(y_p = A e^{5t}\text{.}\)
Step 3. No overlap.
Step 4. Substitute: \(-7A e^{5t} = 3e^{5t} → A=-\frac{3}{7}\text{.}\)
Step 5. General: \(y=c_1 e^{6t} + c_2 e^{-2t} - \frac{3}{7}e^{5t}\text{.}\)
Step 6. Apply ICs:
\begin{gather*}
c_1 + c_2 = 3\\
6c_1 - 2c_2 = 2
\end{gather*}
Solve → \(c_1=1, c_2=2\text{.}\) \(y(t) = e^{6t} + 2e^{-2t} - \frac{3}{7}e^{5t}\text{.}\)
10.4.21.
10.4.22.
11 Laplace Transforms
11.6 Exercises
🏋️♂️ Practice Drills
11.6.9.
Solution.
We will use properties in the table as follows.
\begin{align*}
Y(s) \amp = \lap{ y(t) }\\
\amp = \lap{ 15 - 4e^{9t} + 11t^3 }\\
\amp = 15\lap{ 1 } - 4\lap{ e^{9t} } + 11\lap{ t^3 } \qquad (\knowl{./knowl/xref/lt-P1.html}{\text{P\(_1\)}})\\
\amp =
{\color{blue}\us{s \gt 0}{{\ub{{\color{black}15\cdot \dfrac{1}{s}}}}}\
{\color{black}-\ } \us{s \gt 9}{\ub{{\color{black}4\cdot \dfrac{1}{s - 9}}}}\
{\color{black}+\ } \us{s \gt 0}{\ub{{\color{black}11 \cdot \dfrac{3!}{s^{3 + 1}}}}}} \qquad
( \knowl{./knowl/xref/lt-L1-table.html}{\text{\(L1\)}},
\knowl{./knowl/xref/lt-L2-table.html}{\text{\(L2\)}},
\knowl{./knowl/xref/lt-L3-table.html}{\text{\(L3\)}})\\
\amp = \dfrac{15}{s} - \dfrac{4}{s-9} + \dfrac{66}{s^4}, \hspace{0.5cm} s \gt 9
\end{align*}
11.6.10.
Solution.
We will use properties in the table as follows.
\begin{align*}
\amp = \lap{ e^{3t}\sin(6t) - t^3e^{-5t} }\\
\amp = \lap{ e^{3t}\sin(6t) } - \lap{ t^3e^{-5t} } \qquad (\knowl{./knowl/xref/lt-P1.html}{\text{P\(_1\)}}) \\
\amp = \ub{\dfrac{6}{(s-3)^2 + 6^2}}_{s \gt 3} - \ub{\dfrac{3!}{\Big(s - (-5)\Big)^{3+1}}}_{s >-5} \qquad (\knowl{./knowl/xref/lt-L7-table.html}{\text{\(L7\)}}, \knowl{./knowl/xref/lt-L6-table.html}{\text{\(L6\)}}) \\
\amp = \dfrac{6}{(s-3)^2 + 36} - \dfrac{6}{(s+5)^4}, \hspace{0.5cm} s \gt 3
\end{align*}
11.6.11.
Solution.
Before we begin, we note that it’s very tempting to think that because we know the Laplace transforms of both \(t^2\) and \(\cos(8t),\) we can simply multiply those together to get the desired Laplace transform. However, this is not the case, just as similar statements were not true for finding the derivatives and integrals of the products of functions. Rather, we will need to use property \(L13\), with \(n = 2\) and \(f(t) = \cos(8t).\)
\begin{align*}
G(s) \amp = \lap{ g(t) }\\
\amp = \lap{ t^2 \cos(8t) }\\
\amp = \lap{ t^2 f(t) }\\
\amp = (-1)^2\cdot \dfrac{d^2}{ds^2}\big( F(s) \big)
\end{align*}
We need to know what \(F(s)\) is before we can proceed. Let’s go back to the naming system we have instituted. If we have a capital \(F(s),\) that is the Laplace transform of a function lower case \(f(t).\) We identified that function previously: \(f(t) = \cos(8t).\) We use \(L5\) to find its Laplace transform.
\begin{equation*}
F(s) = \dfrac{s}{s^2 + 64}, s >0
\end{equation*}
Then we continue finding \(G(s)\) by taking two derivatives (using the quotient rule for derivatives; details are omitted here).
\begin{align*}
G(s) \amp = (-1)^2\cdot \dfrac{d^2}{ds^2}\big( F(s) \big)\\
\amp = 1 \cdot \dfrac{d^2}{ds^2}\left( \dfrac{s}{s^2 + 64} \right)\\
\amp = \dfrac{d}{ds}\left( \dfrac{-s^2 + 64}{(s^2 + 64)^2} \right)\\
\amp = \dfrac{2s(s^2 - 192)}{(s^2 + 64)^3}
\end{align*}
11.6.13.
11.6.14.
11.6.15.
11.6.16.
Solution.
\begin{align*}
\lap{e^{2t} - t^3 - \sin (5t)}
\amp = \lap{ e^{2t} } - \lap{ t^3 } - \lap{ \sin(5t) } \quad \knowl{./knowl/xref/lt-P1.html}{\text{P\(_1\)}}\\
\amp = \dfrac{1}{s-2} - \dfrac{3!}{s^{3+1}} - \dfrac{5}{s^2 + 5^2} \quad
\knowl{./knowl/xref/lt-L2-table.html}{\text{L\(_2\)}}, \knowl{./knowl/xref/lt-L3-table.html}{\text{L\(_3\)}}, \knowl{./knowl/xref/lt-L4-table.html}{\text{L\(_4\)}}\\
\amp = \dfrac{1}{s-2} - \dfrac{6}{s^4} - \dfrac{5}{s^2 + 25}
\end{align*}
11.6.17.
Solution.
\begin{align*}
F(s) \amp = \lap{ f(t) } \\
\amp = \lap{ e^{-2t}\sin(2t) + t^2 e^{3t} } \\
\amp = \lap{ e^{-2t}\sin(2t)} + \lap{t^2 e^{3t} } \\
\amp = \dfrac{2}{\big(s- (-2) \big)^2 + 2^2} + \dfrac{2!}{(s-3)^{2+1}} \mbox{(\knowl{./knowl/xref/lt-L8-table.html}{\text{L\(_8\)}}, \knowl{./knowl/xref/lt-L7-table.html}{\text{L\(_7\)}})} \\
\amp = \dfrac{2}{(s+2)^2 + 4} + \dfrac{2}{(s-3)^3}
\end{align*}
11.6.18.
Solution.
For this solution, we will need to use property L\(_{11}\) with \(\ds f(t) = \cos(6t).\) We will need to know the Laplace transform for this function, so let’s do that now.
\begin{align*}
F(s) \amp = \lap{ f(t) } \\
\amp = \lap{ \cos(6t) } \\
\amp = \dfrac{s}{s^2 + 6^2} \mbox{(\knowl{./knowl/xref/lt-L5-table.html}{\text{L\(_5\)}})} \\
\amp = \dfrac{s}{s^2 + 36}
\end{align*}
Then we have the following. Note that when we use the quotient rule to take the derivative of \(\ds F(s)\text{.}\)
\begin{align*}
Q(s) \amp = \lap{ q(t) } \\
\amp = \lap{ 8t\cos(6t) + e^{3t}\sin(4t) } \\
\amp = 8\lap{ t\cos(6t) } + \lap{ e^{3t}\sin(4t) } \\
\amp = 8\lap{ t\cos(6t) } + \dfrac{4}{(s-3)^2 + 4^2} \mbox{(\knowl{./knowl/xref/lt-L8-table.html}{\text{L\(_8\)}})} \\
\amp = 8\lap{ t\cos(6t) } + \dfrac{4}{(s-3)^2 + 16} \\
\amp = 8\lap{ t\cdot f(t) } + \dfrac{4}{(s-3)^2 + 16} \\
\amp = 8\cdot (-1)^1 \cdot \dfrac{d}{ds}F(s) + \dfrac{4}{(s-3)^2 + 16} \mbox{(\knowl{./knowl/xref/lt-L11.html}{\text{L\(_{11}\)}})} \\
\amp = -8 \cdot \dfrac{d}{ds}\left( \dfrac{s}{s^2 + 36} \right) + \dfrac{4}{(s-3)^2 + 16} \\
\amp = -8 \cdot \dfrac{(s^2 + 36)\cdot 1 - s\cdot (2s + 0)}{(s^2 + 36)^2} + \dfrac{4}{(s-3)^2 + 16} \\
\amp = -8 \cdot \dfrac{s^2 + 36 - 2s^2}{(s^2 + 36)^2} + \dfrac{4}{(s-3)^2 + 16} \\
\amp = -\dfrac{8(36 - s^2)}{(s^2 + 36)^2} + \dfrac{4}{(s-3)^2 + 16}
\end{align*}
11.6.19.
Solution.
For this solution, we will need to use L\(_{11}\) with \(f(t) = \sin(3t)\text{.}\) We will need the Laplace transform of this function, so let’s compute it now.
\begin{align*}
F(s) \amp = \lap{ f(t) } \\
\amp = \lap{ \sin(3t) } \\
\amp = \dfrac{3}{s^2 + 3^2} \mbox{(\knowl{./knowl/xref/lt-L5-table.html}{\text{L\(_5\)}})} \\
\amp = \dfrac{3}{s^2 + 9}
\end{align*}
In using L\(_{11}\), we will also need the second derivative of \(\ds F(s),\text{,}\) so we compute it now. Note that we will use the chain rule when we take the derivative of \(\ds (s^2 + 9)^2\text{.}\)
\begin{align*}
F'(s) \amp = \dfrac{d}{ds}\left( \dfrac{3}{s^2 + 9} \right) \\
\amp = \dfrac{(s^2 + 9) \cdot 0 - 3 \cdot (2s+0)}{(s^2 + 9)^2} \\
\amp = \dfrac{-6s}{(s^2 + 9)^2} \\
F''(s) \amp = \dfrac{(s^2 + 9)^2 \cdot (-6) - (-6s) \cdot 2(s^2 + 9)(2s)}{(s^2 + 9)^4} \\
\amp = \dfrac{-6(s^2 + 9)^2 + 24s^2(s^2 +9)}{(s^2 + 9)^4} \\
\amp = \dfrac{6(s^2 + 9)\left[-(s^2 + 9) + 4s^2\right]}{(s^2 + 9)^4} \\
\amp = \dfrac{6\left[-s^2 - 9 + 4s^2\right]}{(s^2 + 9)^3} \\
\amp = \dfrac{6\left[3s^2 - 9\right]}{(s^2 + 9)^3} \\
\amp = \dfrac{18\left[s^2 - 3\right]}{(s^2 + 9)^3}
\end{align*}
Then we have the following.
\begin{align*}
P(s) \amp = \lap{ t^2 \sin(3t) } \\
\amp = \lap{ t^2 f(t) } \\
\amp = (-1)^2 \cdot F''(s) \mbox{(\knowl{./knowl/xref/lt-L11.html}{\text{L\(_{11}\)}})} \\
\amp = 1 \cdot \dfrac{18\left[s^2 - 3\right]}{(s^2 + 9)^3} \\
\amp = \dfrac{18\left[s^2 - 3\right]}{(s^2 + 9)^3}
\end{align*}
✍🏻 Problems
11.6.1.
Solution.
Let’s use the definition:
\begin{align*}
\lap{11t} \amp = \int_0^{\infty} \left(11t\right) e^{-st}\ dt\\
\amp = 11\lim_{b \to \infty} \ub{\int_0^b te^{-st} dt}_{\large I}
\end{align*}
Note: \(s\) cannot be \(0\).
When \(s=0\text{,}\) the integral becomes
\begin{align*}
\amp
= \lim_{b \to \infty} \int_0^b t\ dt
= \lim_{b \to \infty} \dfrac{t^2}{2}\Big|_0^b
= \dfrac{1}{2} \lim_{b \to \infty} b^2 = \infty
\end{align*}
Therefore, we must have \(s\ne 0\) for this integral to be finite.
\begin{align*}
\lap{11t}
\amp = 11\lim_{b \to \infty} \left[-\dfrac{b}{s}e^{-sb} - \dfrac{1}{s^2}e^{-sb} + \dfrac{1}{s^2}\right]
\end{align*}
\begin{align*}
\amp = 11 \left[-\lim_{b \to \infty}\dfrac{b}{s}e^{-sb} - \lim_{b \to \infty}\dfrac{1}{s^2}e^{-sb} + \lim_{b \to \infty}\dfrac{1}{s^2}\right]\\
\amp = 11 \left[-\dfrac{1}{s}\lim_{b \to \infty}be^{-sb} - \dfrac{1}{s^2}\lim_{b \to \infty}e^{-sb} + \dfrac{1}{s^2}\right]
\end{align*}
As long as \(s \gt 0\text{,}\) the limit becomes:.
If \(s \lt 0\text{,}\) then as \(b \to \infty\text{,}\) we would have \(e^{-sb} \to \infty\) and so
\begin{equation*}
\lim_{b \to \infty} \os{\infty}{\os{\uparrow}{\boxed{b}}}\ \us{\infty}{\us{\downarrow}{\boxed{e^{-sb}}}} = \infty.
\end{equation*}
This shows the Laplace transform would not exist if \(s \lt 0\text{.}\) Therefore, we must require \(s \gt 0\text{.}\)
\begin{align*}
\lap{11t}
\amp = 11 \left[-\dfrac{1}{s}\cdot 0 - \dfrac{1}{s^2}\cdot 0 + \dfrac{1}{s^2}\right]\\
\amp = \dfrac{11}{s^2}, \quad s \gt 0
\end{align*}
Hence, we have \(\lap{ 11t } = \dfrac{11}{s^2}\text{,}\) under the condition that \(s > 0\text{.}\)
11.6.2.
Solution.
\begin{align*}
\lap{7t + e^{5t}} \amp = \int_0^{\infty}e^{-st}\cdot (7t + e^{5t}) dt \\
\amp = \int_0^{\infty}\left(e^{-st}\cdot 7t + e^{-st}\cdot e^{5t}\right) dt \\
\amp = \int_0^{\infty}e^{-st}\cdot 7t\ dt + \int_0^{\infty}e^{-st}\cdot e^{5t}\ dt \\
\amp = \lap{ 7t } + \lap{ e^{5t} } \\
\amp = \dfrac{7}{s^2} + \dfrac{1}{s-5}
\end{align*}
11.6.3.
11.6.4.
Solution.
To compute this, we use the definition of the Laplace transform.
\begin{align*}
\lap{-40e^{3t}} \amp = \int_0^{\infty} e^{-st} \cdot (-40e^{3t}) \ dt \\
\amp = -40 \int_0^{\infty} e^{-st+3t} \ dt \\
\amp = -40 \int_0^{\infty} e^{-(s-3)t} \ dt \\
\amp = -40 \left[ \dfrac{1}{s-3} e^{-(s-3)t} \right]_0^{\infty} \\
\amp = -40 \left( \lim_{t \to \infty} \dfrac{1}{s-3} e^{-(s-3)t} - \dfrac{1}{s-3} \right) \\
\amp = -40 \left( 0 - \dfrac{1}{s-3} \right) \\
\amp = \dfrac{40}{s-3}
\end{align*}
11.6.5.
Solution.
By the above definition, we have
\begin{align*}
\lap{ 15 } \amp = \int_0^{\infty} \left(15\right)e^{-st}dt \qquad \text{(improper integral)} \\
\amp = \lim_{b \to \infty}\int_0^b 15 e^{-st}dt \\
\amp = 15 \lim_{b \to \infty}\int_0^b e^{-st}dt \qquad (s\text{ is constant here})
\end{align*}
At this point, we need to assume that \(s\ne0\text{.}\).
Next, we integrate using the substitution \(u =- st \text{.}\)
\begin{align*}
\amp = 15\lim_{b \to \infty} -\dfrac{1}{s}e^{-st}\Big|_{t=0}^{t=b}\\
\amp = 15\lim_{b \to \infty} -\dfrac{1}{s}\left[ e^{-sb} - e^{-s\cdot 0} \right]\\
\amp = 15 \cdot -\dfrac{1}{s} \lim_{b \to \infty} \Big[ e^{-sb} - 1 \Big]\\
\amp = -\dfrac{15}{s} \Big[ \lim_{b \to \infty} e^{-sb} - \us{=\ 1}{\ub{\lim_{b \to \infty} 1}} \Big] \qquad(s > 0)\\
\amp = -\dfrac{15}{s} \left[0 - 1\right] = \dfrac{15}{s}
\end{align*}
11.6.6.
Solution.
We use the definition of the Laplace transform to get us started.
\begin{align*}
\lap{ e^{7t} } \amp = \int_0^{\infty} \left(e^{7t}\right)e^{-st}\ dt \\
\amp = \lim_{b \to \infty}\int_0^b e^{-st + 7t}\ dt \\
\amp = \lim_{b \to \infty}\int_0^b e^{(7-s)t}\ dt
\end{align*}
As before, we need to restrict some values of \(s\) in order for this improper integral to exist. In this case, we will need \(7 - s\text{,}\) in the exponent, to be non-zero and negative. That is, we need
\begin{equation*}
7 - s \lt 0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}7 \lt s.
\end{equation*}
\begin{align*}
\amp = \lim_{b \to \infty}\int_0^b e^{(7-s)t}dt \\
\amp = \lim_{b \to \infty} \dfrac{1}{7-s}e^{(7-s)t}\Bigg|_0^b\\
\amp = \dfrac{1}{7-s}\lim_{b \to \infty} \left[ e^{(7-s)b} - e^{(7-s)\cdot 0}\right]\\
\amp = \dfrac{1}{7-s}\lim_{b \to \infty} \left[ e^{(7-s)b} - 1\right]\\
\amp = \dfrac{1}{7-s} \left[\lim_{b \to \infty}e^{(7-s)b} - \lim_{b \to \infty}1\right]\\
\amp = \dfrac{1}{7-s} \left[ 0 - 1\right]\\
\amp = \dfrac{-1}{7-s}\\
\amp = \dfrac{1}{s - 7},\quad s \gt 7
\end{align*}
Thus,
\begin{equation*}
\ds \lap{e^{7t}} = \dfrac{1}{s - 7}\text{.}
\end{equation*}
11.6.7.
Solution.
We start by applying the definition of the Laplace transform:
\begin{equation*}
\lap{ \cos(3t)} = \int_0^{\infty} e^{-st} \cdot \cos(3t)\ dt\text{.}
\end{equation*}
Rather than directly integrating, we will use a modified Euler’s Formula to express cosine in terms of \(e\)
\begin{equation*}
\cos(3t) = \dfrac12\left(e^{3it} + e^{-3it}\right).
\end{equation*}
Substituting this into the integral gives:
\begin{align*}
\lap{ \cos(3t)}
\amp = \dfrac12 \int_0^{\infty} e^{-st} \left(e^{3it} + e^{-3it}\right)\ dt\\
\amp = \dfrac12\left[\int_0^{\infty} e^{-st}\cdot e^{3it}\ dt + \int_0^{\infty} e^{-st}\cdot e^{-3it}\ dt\right]\\
\amp = \dfrac{1}{2} \left[\lap{e^{3it}} + \lap{e^{-3it}}\right]\\
\amp = \dfrac{1}{2} \left[\dfrac{1}{s - 3i} + \dfrac{1}{s + 3i}\right] \qquad (\text{by } \knowl{./knowl/xref/lt-L2-table.html}{\text{L\(_2\)}})\\
\amp = \dfrac{1}{2} \left[\dfrac{s + 3i + s - 3i}{(s - 3i)(s + 3i)}\right]\\
\amp = \dfrac{s}{s^2 + 9}.
\end{align*}
Therefore, the Laplace transform of \(\cos(3t)\) is:
\begin{equation*}
\lap{\cos(3t)} = \dfrac{s}{s^2 + 9}.
\end{equation*}
11.6.8.
Solution.
As with cosine, we begin with the definition of the Laplace transform,
\begin{equation*}
\lap{ \sin(-4t)} = \int_0^{\infty} e^{-st} \cdot \sin(-4t)\ dt
\end{equation*}
and rewrite sine using Euler’s formula,
\begin{equation*}
\sin(-4t) = \dfrac{e^{-4it} - e^{4it}}{2i}.
\end{equation*}
Substituting this into the integral, we get:
\begin{align*}
\lap{ \sin(-4t)}
=\amp \dfrac{1}{2i} \int_0^{\infty} e^{-st} \left(e^{-4it} - e^{4it}\right)\, dt\\
=\amp \dfrac{1}{2i} \left[\int_0^{\infty} e^{-(s + 4i)t}\, dt - \int_0^{\infty} e^{-(s - 4i)t}\, dt\right]\\
=\amp \dfrac{1}{2i} \left[\lap{e^{-4it}} - \lap{e^{4it}}\right].\\
=\amp \dfrac{1}{2i} \left[\dfrac{1}{s + 4i} - \dfrac{1}{s - 4i}\right] \qquad (\text{by } \knowl{./knowl/xref/lt-L2-table.html}{\text{L\(_2\)}})\\
=\amp \dfrac{1}{2i} \left[\dfrac{s - 4i - (s + 4i)}{(s + 4i)(s - 4i)}\right]\\
=\amp -\dfrac{4}{s^2 + 16}.
\end{align*}
Thus, the Laplace transform of \(\sin(-4t)\) is:
\begin{equation*}
\lap{\sin(-4t)} = -\dfrac{4}{s^2 + 16}.
\end{equation*}
12 Laplace Transform Method
12.5 Exercises
🏋️♂️ Practice Drills
12.5.2. Forward Transform the Equation.
Solution.
Applying the Laplace Transform to both sides:
\begin{equation*}
\lap{y''} - 4\lap{y'} + 6\lap{y} = \lap{e^{2t}}
\end{equation*}
\begin{gather*}
s^2Y(s) - s\cdot y(0) - y'(0) - 4(sY(s) - y(0)) + 6Y(s)
= \dfrac{1}{s-2}\\
s^2Y(s) - s - 4(sY(s) - 1) + 6Y(s)
= \dfrac{1}{s-2}\\
(s^2 - 4s + 6)Y(s) = \dfrac{1}{s-2} + s + 4
\end{gather*}
12.5.31.
12.5.32.
Solution.
This expression can be split into two parts: \(\dfrac{2(s + 3)}{(s + 3)^2 + 4} \) and \(\dfrac{7}{(s + 3)^2 + 4} \text{.}\) The first part corresponds to \(e^{-3t} \cdot \cos(2t) \) and the second part corresponds to \(e^{-3t} \cdot \sin(2t) \text{,}\) so the inverse Laplace transform is \(2e^{-3t} \cos(2t) + 7e^{-3t} \sin(2t) \text{.}\)
12.5.33.
12.5.34.
12.5.35.
12.5.36.
Solution.
The discriminant of the denominator \(s^2 - 6s + 14\) is:
\begin{equation*}
b^2 - 4ac = (-6)^2 - 4(1)(14) = 36 - 56 = -20\text{,}
\end{equation*}
indicating that completing the square is necessary:
\begin{align*}
s^2 - 6s + 14 = (s^2 - 6s) + 14 =\amp (\ob{s^2 - 6s + 9}^{(s - 3)^2} - 9) + 14 \\
=\amp (s - 3)^2 + 5 \text{.}
\end{align*}
Rewriting \(K(s)\) as:
\begin{equation*}
\dfrac{11}{s^2 - 6s + 14}
= \ub{\dfrac{11}{(s - 3)^2 + 5}}_{\large \text{Match with } L_7}
= \dfrac{11}{\sqrt{5}}\ub{\dfrac{\sqrt{5}}{(s - 3)^2 + 5}}_{\large \knowl{./knowl/xref/lt-L7-table.html}{\text{\(L_7\)}} (a = 3, b = \sqrt{5})}\text{.}
\end{equation*}
Therefore,
\begin{align*}
k(t) = \ilap{K(s)} \amp = \dfrac{11}{\sqrt{5}}e^{3t}\sin(\sqrt{5}t)\text{.}
\end{align*}
12.5.37.
Solution.
Completing the square for the denominator of \(P(s)\) gives:
\begin{equation*}
\dfrac{s+3}{s^2 + 2s + 10} = \dfrac{s+3}{(s + 1)^2 + 9}\text{.}
\end{equation*}
However, the numerator \(s + 3\) does not match \(s + 1\text{.}\) To resolve this, we rewrite \(3\) as \(1 + 2\) and group terms:
\begin{align*}
\dfrac{s+3}{s^2 + 2s + 10}
\amp = \dfrac{s + 1 + 2}{(s + 1)^2 + 9}\\
\amp = \dfrac{s+1}{(s + 1)^2 + 9} + \dfrac{2}{(s + 1)^2 + 9}\\
\amp = \ub{\dfrac{s+1}{(s + 1)^2 + 9}}_{\large \knowl{./knowl/xref/lt-L8-table.html}{\text{L\(_8\)}}\ (a=-1, b=3)}
+ \dfrac{2}{3}\ub{\dfrac{3}{(s + 1)^2 + 9}}_{\large \knowl{./knowl/xref/lt-L7-table.html}{\text{L\(_7\)}}\ (a=-1, b=3)}\text{,}
\end{align*}
Now, apply the inverse Laplace transform:
\begin{align*}
p(t) \amp = \ilap{P(s)}\\
\amp = \ilap{ \dfrac{s+1}{(s + 1)^2 + 9} } + \dfrac{2}{3}\ilap{ \dfrac{3}{(s + 1)^2 + 9} }\\
\amp = e^{-t}\cos(3t) + \dfrac{2}{3}e^{-t}\sin(3t)
\end{align*}
12.5.38.
Solution.
First, factor the quadratic denominator:
\begin{equation*}
s^2 - s - 6 = (s + 2)(s - 3).
\end{equation*}
Now decompose \(P(s)\) into partial fractions:
\begin{equation*}
\dfrac{s}{(s + 2)(s - 3)} = \dfrac{A}{s + 2} + \dfrac{B}{s - 3}.
\end{equation*}
Multiply both sides by \((s + 2)(s - 3)\text{,}\)
\begin{equation*}
s = A(s - 3) + B(s + 2).
\end{equation*}
and solve for \(A\) and \(B\) by selecting convenient values for \(s\text{:}\) The partial fraction decomposition is:
\begin{equation*}
\dfrac{s}{(s + 2)(s - 3)} = \dfrac{2/5}{s + 2} + \dfrac{3/5}{s - 3}.
\end{equation*}
Therefore:
\begin{equation*}
\ilap{P(s)} = \dfrac25 e^{-2t} + \dfrac35 e^{3t}.
\end{equation*}
12.5.39.
12.5.40.
Solution.
Decompose into partial fractions:
\begin{equation*}
Y(s) = \dfrac{A}{s + 1} + \dfrac{B}{s + 3}\text{.}
\end{equation*}
Find \(A\) and \(B\text{:}\)
\begin{equation*}
A = 1, B = \dfrac{1}{2}\text{.}
\end{equation*}
Use the inverse Laplace transform:
\begin{equation*}
y(t) = e^{-t} + \dfrac{1}{2}e^{-3t}\text{.}
\end{equation*}
12.5.41.
Solution.
Decompose into partial fractions:
\begin{equation*}
Y(s) = \dfrac{A}{s} + \dfrac{B}{s + 4}\text{.}
\end{equation*}
Find \(A\) and \(B\text{:}\)
\begin{equation*}
A = \dfrac{5}{4}, B = -\dfrac{5}{4}\text{.}
\end{equation*}
Use the inverse Laplace transform:
\begin{equation*}
f(t) = \dfrac{5}{4}(1 - e^{-4t})\text{.}
\end{equation*}
12.5.42.
Solution.
As in the previous examples, the denominator is a second-degree polynomial; therefore, it is sensible for us to begin by completing the square in the denominator as we did in the previous two examples.
\begin{align*}
\dfrac{s+9}{s^2 - 2s - 3} \amp = \dfrac{s+9}{(s^2 - 2s) - 3}\\
\amp = \dfrac{s+9}{(s^2 - 2s + 1) - 3 - 1}\\
\amp = \dfrac{s+9}{(s-1)^2 - 4}\\
\amp = \dfrac{s+9}{(s-1)^2 - 2^2}.
\end{align*}
Take a careful look at the denominator here. It’s really close to matching \(L7\) or \(L8\), but it is not a match because of the negative sign in front of the \(2^2.\) We need to change course when this happens. Another algebraic manipulation that we might consider is a partial fraction decomposition.
📝
We revert to the original expression, but this time, instead of completing the square, we factor the denominator.
\begin{equation*}
\dfrac{s+9}{s^2 - 2s - 3} = \dfrac{s+9}{(s-3)(s+1)}.
\end{equation*}
Since each of the factors in the denominator is a distinct linear factor, we know that the form of the partial fraction decomposition is
\begin{equation*}
\dfrac{s+9}{(s-3)(s+1)} = \dfrac{A}{s-3} + \dfrac{B}{s+1}.
\end{equation*}
Our next goal is to determine the coefficients \(A\) and \(B\) in these equations. There are multiple ways to achieve this, and we demonstrate just one here. We multiply both sides of the equation by the least common denominator, \((s-3)(s+1)\text{,}\) and then expand and collect like-terms, as shown.
\begin{align*}
(s-3)(s+1) \cdot \dfrac{s+9}{(s-3)(s+1)} \amp = \dfrac{A}{s-3}\cdot (s-3)(s+1) + \dfrac{B}{s+1} \cdot (s-3)(s+1)\\
s+9 \amp = A(s+1) + B(s-3)\\
s+9 \amp = As + A + Bs - 3B\\
s+9 \amp = (As + Bs) + (A - 3B)\\
1s+9 \amp = (A+B)s + (A - 3B).
\end{align*}
At this point, we have a polynomial on the left hand side and a polynomial on the right hand side. The only way these can be equal is if the corresponding coefficients are equal. That is, the coefficient on \(s\) on the left-hand side is 1, while the coefficient on \(s\) on the right-hand side of the equation is \(A+B\text{.}\) Since the polynomials are equal, they are equal. That is, \(A + B = 1.\) Similarly, if we equate the constants, we have \(A - 3B = 9.\) Thus, we have the following system of two linear equations in terms of two unknown variables, \(A\) and \(B\text{.}\)
\begin{align*}
A + B \amp = 1\\
A - 3B \amp = 9.
\end{align*}
There are many ways to solve such an equation, and you are encouraged to choose the solution technique you like the most. Here we will solve the first equation for \(A\text{,}\) and then substitute into the second equation,
\begin{align*}
A + B \amp = 1 \amp A - 3B \amp = 9\\
A \amp = 1 - B \amp \amp\\
\amp \amp (1 - B) - 3B \amp = 9\\
\amp \amp 1 - 4B \amp = 9\\
\amp \amp -4B \amp = 8\\
\amp \amp B \amp = -2\\
A \amp = 1 - (-2) \amp \amp\\
\amp = 3 \amp \amp
\end{align*}
hence we have
\begin{equation*}
\dfrac{s+9}{(s-3)(s+1)} = \dfrac{3}{s-3} + \dfrac{-2}{s+1}.
\end{equation*}
Remember that our goal is to take the inverse Laplace transform. Our algebraic manipulation was helpful because we took a more complex expression and rewrote it as two simpler fractions. We can now use \(L2\) to find the inverse Laplace transform as follows.
\begin{align*}
\ilap{ \dfrac{s+9}{s^2 - 2s - 3} } \amp = \ilap{ \dfrac{3}{s-3} + \dfrac{-2}{s+1} }\\
\amp = 3 \ilap{ \dfrac{1}{s-3} } - 2 \ilap{ \dfrac{1}{s+1} } \knowl{./knowl/xref/lt-R1-table.html}{\text{R\(_1\)}} \\
\amp = 3e^{3t} - 2e^{-t} \quad \text{ by } \knowl{./knowl/xref/lt-L2-table.html}{\text{L2}}
\end{align*}
12.5.43.
Solution.
📝
\begin{equation*}
s^2 - 4s + 29 = (s-2)^2 + 25.
\end{equation*}
(If you still don’t remember how to complete the square, look up that primer and do the previous exercises in that section above.)
Let’s rewrite the given expression as follows.
\begin{align*}
\dfrac{s-6}{s^2 - 4s + 29} \amp = \dfrac{s-6}{(s-2)^2 + 25} \\
\amp = \dfrac{s-6}{(s-2)^2 + 5^2}
\end{align*}
We’ve got the denominator in exactly the right form; it looks just like \((s-a)^2 + b^2,\) with \(a = 2\) and \(b = 5\text{.}\) As in the previous section, once we’ve gotten the denominator in shape, we turn our attention to the numerator. If we look back at the two forms we are trying to match, we see that our expression has an \(s\) in the numerator, so it’s more like \(\dfrac{s-a}{(s-a)^2 + b^2}\text{.}\) It would be exactly right if we had \(s-a\) in the numerator, which in this case would be \(s-2\text{.}\)
What we do have in the numerator is \(s-6\text{;}\) and we would like it to be \(s-2,\) which means if we added 4, we’d have exactly the right thing. If we want to add 4, we’ll need to compensate by also subtracting 4, like this:
\begin{align*}
\dfrac{s-6}{(s-2)^2 + 5^2}\amp = \dfrac{s-6+4-4}{(s-2)^2 + 5^2}, \\
\amp = \dfrac{(s-6+4)-4}{(s-2)^2 + 5^2}, \\
\amp = \dfrac{(s-2) - 4}{(s-2)^2 + 5^2}.
\end{align*}
Great! Now we can split this single fraction into two separate fractions:
\begin{equation*}
\dfrac{(s-2) - 4}{(s-2)^2 + 5^2}
= \underbrace{\dfrac{s-2}{(s-2)^2 + 5^2}}_{match!} - \dfrac{4}{(s-2)^2 + 5^2}
\end{equation*}
We’re almost there! The first fraction is a perfect match for the form \(\dfrac{s-a}{(s-a)^2 + b^2}\) (with \(a = 2\) and \(b = 5\)); but we still have another expression that isn’t a match yet. The remaining fraction looks like it could eventually match the form \(\dfrac{b}{(s-a)^2 + b^2}\text{.}\) We need a 5 in the numerator, but we currently have a 4. But we can fix that as we did in the previous section:
\begin{align*}
\dfrac{4}{(s-2)^2 + 5^2} \amp = 4\cdot\dfrac{1}{(s-2)^2 + 5^2} \\
\amp = 4\cdot\dfrac{5}{5}\cdot
\dfrac{1}{(s-2)^2 + 5^2} \\
\amp = 4\cdot\dfrac{1}{5}\cdot
\dfrac{5}{(s-2)^2 + 5^2} \\
\amp = \dfrac{4}{5}\cdot
\dfrac{5}{(s-2)^2 + 5^2}
\end{align*}
Now let’s put it all together. Here’s everything we did:
\begin{align*}
\dfrac{s-6}{s^2 - 4s + 9} \amp = \dfrac{s-6}{(s-2)^2 + 25} \mbox{(we completed the square)}\\
\amp = \dfrac{s-6}{(s-2)^2 + 5^2} \mbox{(we rewrote the denominator)}\\
\amp = \dfrac{s-6+4-4}{(s-2)^2 + 5^2} \mbox{(we added and subtracted 4 in the numerator...)}\\
\amp = \dfrac{(s-2) - 4}{(s-2)^2 + 5^2} \mbox{(...so that we could make the numerator match part of the denom)}\\
\amp = \dfrac{s-2}{(s-2)^2 + 5^2} - \dfrac{4}{(s-2)^2 + 5^2} \mbox{(split into fractions)}\\
\amp = \dfrac{s-2}{(s-2)^2 + 5^2} - 4\cdot\dfrac{5}{5}\cdot \dfrac{1}{(s-2)^2 + 5^2} \mbox{(work on the second fraction...)}\\
\amp = \dfrac{s-2}{(s-2)^2 + 5^2} - \dfrac{4}{5}\cdot \dfrac{5}{(s-2)^2 + 5^2}\mbox{(...so now it is also in form)}
\end{align*}
As mentioned before, being able to use appropriate algebra to "match" forms is really important when we work with Laplace Transforms. Since it’s really just algebra, now is a great time to practice that skill, so when we are in the middle of studying Laplace Transforms, you can just focus on the "new" stuff.
12.5.44.
12.5.45.
12.5.46.
12.5.47.
12.5.48.
12.5.49.
12.5.50.
12.5.51.
Solution.
Note that as the quadratic term in the denominator does not factor, the denominator contains an irreducible quadratic factor and a repeated linear factor. We’ll proceed by simplifying this complicated fraction with a Partial Fraction Decomposition of the form
\begin{equation*}
\dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}
= \dfrac{As + B}{s^2 - 6s + 25} + \dfrac{C}{s+1} + \dfrac{D}{(s+1)^2}.
\end{equation*}
You may also consider using technology to find a partial fraction decomposition. You should get
\begin{equation*}
\dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2} = \dfrac{s + 2}{s^2 - 6s + 25} + \dfrac{3}{s+1} + \dfrac{-2}{(s+1)^2}.
\end{equation*}
Click here for the details.
\begin{align*}
4s^3 \amp - 13s^2 + 74s + 27\\
\amp = (As + B)(s+1)^2 + C(s+1)(s^2 - 6s + 25) + D(s^2 - 6s + 25)\\
\amp = (As + B)(s^2 + 2s + 1) + (Cs+C)(s^2 - 6s + 25) + Ds^2 - 6Ds + 25D\\
\amp = As^3 + 2As^2 + As + Bs^2 + 2Bs + B + Cs^3 - 6Cs^2 + 25Cs + Cs^2 - 6Cs + 25C + Ds^2 - 6Ds + 25D\\
\amp = (A + C)s^3 + (2A+ B - 6C + C + D)s^2 + (A+ 2B + 25C- 6C- 6D)s + (B + 25C + 25D)\\
\amp = (A + C)s^3 + (2A+ B - 5C + D)s^2 + (A+ 2B + 19C- 6D)s + (B + 25C + 25D)
\end{align*}
Equating coefficients gives us four equations in four unknowns.
\begin{align*}
A+C \amp = 4 \amp 2A+B-5C+D \amp = -13 \amp A + 2B + 19C - 6D \amp = 74 \amp B+ 25C + 25D \amp = 27\\
A \amp = 4-C\amp \amp \amp \amp \amp\amp\\
\amp \amp 2(4 - C)+B-5C+D \amp = -13 \amp (4-C)+2B+19C-6D \amp = 74 \amp\amp\\
\amp \amp B - 7C + D \amp = -21 \amp 2B+18C-6D \amp = 70 \amp\amp\\
\amp \amp B \amp = 7C - D - 21 \amp \amp \amp\amp\\
\amp \amp \amp \amp 2(7C - D - 21)+18C-6D \amp = 70 \amp\\
(7C - D - 21) + 25C + 25D \amp = 27\\
\amp \amp \amp \amp 32C-8D \amp = 112 \amp\\
32C + 24D \amp = 48\\
\amp \amp \amp \amp 32C \amp = 8D + 112 \amp\amp\\
\amp \amp \amp \amp \amp \amp\\
(8D + 112) + 24D \amp = 48\\
\amp \amp \amp \amp \amp \amp\\
32D \amp = -64\\
\amp \amp \amp \amp \amp \amp\\
D \amp = -2\\
\amp \amp \amp \amp 32C \amp = 8(-2) + 112 \amp\amp\\
\amp \amp \amp \amp \amp = 96 \amp\amp\\
\amp \amp \amp \amp C \amp = 3 \amp\amp\\
\amp \amp B \amp = 7(3) - (-2) - 21\amp \amp \amp \amp\amp\\
\amp \amp \amp = 2 \amp \amp \amp \amp\amp\\
A \amp = 4 - (3)\amp \amp \amp \amp \amp \amp\amp\\
\amp = 1 \amp \amp \amp \amp \amp \amp\amp
\end{align*}
With the partial fraction decomposition in hand, we are prepared to take the inverse Laplace transform, using the same types of algebraic manipulations demonstrated in the previous examples.
\begin{align*}
r(t) \amp = \ilap{ R(s) }\\
\amp = \ilap{ \dfrac{s + 2}{s^2 - 6s + 25} + \dfrac{3}{s+1}
+ \dfrac{-2}{(s+1)^2} } \\
\amp = \ilap{ \dfrac{s + 2}{s^2 - 6s + 25} }
+ 3\cdot \ilap{ \dfrac{1}{s+1} }
- 2\cdot \ilap{\dfrac{1}{(s+1)^2} } \\
\amp = \ilap{ \dfrac{s + 2}{(s-3)^2 + 4^2} }
+ 3e^{-t} - 2te^{-t}\\
\amp = \ilap{ \dfrac{(s - 3) + 5}{(s-3)^2 + 4^2} }
+ 3e^{-t} - 2te^{-t}\\
\amp = \ilap{ \dfrac{s - 3}{(s-3)^2 + 4^2} }
+ 5 \ilap{ \dfrac{1}{(s-3)^2 + 4^2} }
+ 3e^{-t} - 2te^{-t}\\
\amp = e^{3t}\cos(4t)
+ \dfrac{5}{4}\cdot \ilap{ \dfrac{4}{(s-3)^2 + 4^2} }
+ 3e^{-t} - 2te^{-t}\\
\amp = e^{3t}\cos(4t)
+ \dfrac{5}{4}e^{3t}\sin(4t)
+ 3e^{-t} - 2te^{-t}.
\end{align*}
12.5.54.
Solution.
\begin{align*}
\dfrac{s+3}{s^2 - 6s + 10} \amp = \dfrac{s+3}{(s^2 - 6s) + 10}\\
\amp = \dfrac{s+3}{(s^2 - 6s + 9 - 9) + 10}\\
\amp = \dfrac{s+3}{(s^2 - 6s + 9) - 9 + 10}\\
\amp = \dfrac{s+3}{(s - 3)^2 - 9 + 10}\\
\amp = \dfrac{s+3}{(s - 3)^2 + 1}\\
\amp = \dfrac{s+3}{(s - 3)^2 + 1^2}\\
\amp = \dfrac{(s-3+3)+3}{(s - 3)^2 + 1^2}\\
\amp = \dfrac{(s-3)+3+3}{(s - 3)^2 + 1^2}\\
\amp = \dfrac{(s-3)+6}{(s - 3)^2 + 1^2}\\
\amp = \dfrac{s-3}{(s - 3)^2 + 1^2} + \dfrac{6}{(s - 3)^2 + 1^2}\\
\amp = \dfrac{s-3}{(s - 3)^2 + 1^2} + 6\cdot \dfrac{1}{(s - 3)^2 + 1^2}\\
\amp = \dfrac{s - a}{(s - a)^2 +b^2} + 6\cdot \dfrac{b}{(s - a)^2 +b^2},
\mbox{ where and }
\end{align*}
12.5.55.
Solution.
\begin{align*}
\dfrac{s}{s^2 - 8s + 25} \amp = \dfrac{s}{(s^2 - 8s) + 25}\\
\amp = \dfrac{s}{(s^2 - 8s + 16 - 16) + 25}\\
\amp = \dfrac{s}{(s^2 - 8s + 16) - 16 + 25}\\
\amp = \dfrac{s}{(s - 4)^2 - 16 + 25}\\
\amp = \dfrac{s}{(s - 4)^2 +9}\\
\amp = \dfrac{s}{(s - 4)^2 +3^2}\\
\amp = \dfrac{(s - 4) + 4}{(s - 4)^2 +3^2}\\
\amp = \dfrac{s - 4}{(s - 4)^2 +3^2} + \dfrac{4}{(s - 4)^2 +3^2}\\
\amp = \dfrac{s - 4}{(s - 4)^2 +3^2} + 4\cdot \dfrac{1}{(s - 4)^2 +3^2}\\
\amp = \dfrac{s - 4}{(s - 4)^2 +3^2} + \dfrac{3}{3}\cdot 4\cdot \dfrac{1}{(s - 4)^2 +3^2}\\
\amp = \dfrac{s - 4}{(s - 4)^2 +3^2} + \dfrac{1}{3}\cdot 4\cdot \dfrac{3}{(s - 4)^2 +3^2}\\
\amp = \dfrac{s - 4}{(s - 4)^2 +3^2} + \dfrac{4}{3}\cdot \dfrac{3}{(s - 4)^2 +3^2}\\
\amp = \dfrac{s - a}{(s - a)^2 +b^2} + \dfrac{4}{3}\cdot \dfrac{b}{(s - a)^2 +b^2},
\end{align*}
12.5.56.
Solution.
\begin{align*}
\dfrac{s}{s^2 + 8s + 25} \amp = \dfrac{s}{(s^2 + 8s) + 25}\\
\amp = \dfrac{s}{(s^2 + 8s + 16 - 16) + 25}\\
\amp = \dfrac{s}{(s^2 + 8s + 16) - 16 + 25}\\
\amp = \dfrac{s}{(s + 4)^2 - 16 + 25}\\
\amp = \dfrac{s}{(s + 4)^2 +9}\\
\amp = \dfrac{s}{(s + 4)^2 +3^2}\\
\amp = \dfrac{(s + 4) - 4}{(s + 4)^2 +3^2}\\
\amp = \dfrac{s + 4}{(s + 4)^2 +3^2} - \dfrac{4}{(s + 4)^2 +3^2}\\
\amp = \dfrac{s + 4}{(s + 4)^2 +3^2} - 4\cdot \dfrac{1}{(s + 4)^2 +3^2}\\
\amp = \dfrac{s + 4}{(s + 4)^2 +3^2} - \dfrac{3}{3}\cdot 4\cdot \dfrac{1}{(s + 4)^2 +3^2}\\
\amp = \dfrac{s + 4}{(s + 4)^2 +3^2} - \dfrac{1}{3}\cdot 4\cdot \dfrac{3}{(s + 4)^2 +3^2}\\
\amp = \dfrac{s + 4}{(s + 4)^2 +3^2} - \dfrac{4}{3}\cdot \dfrac{3}{(s + 4)^2 +3^2}\\
\amp = \dfrac{s - a}{(s - a)^2 +b^2} - \dfrac{4}{3}\cdot \dfrac{b}{(s - a)^2 +b^2},
\end{align*}
12.5.57.
Solution.
\begin{align*}
\dfrac{s-5}{s^2 +12s + 72} \amp = \dfrac{s-5}{(s^2 +12s) + 72}\\
\amp = \dfrac{s-5}{(s^2 +12s + 36 - 36) + 72}\\
\amp = \dfrac{s-5}{(s^2 +12s + 36) - 36 + 72}\\
\amp = \dfrac{s-5}{(s + 6)^2 - 36 + 72}\\
\amp = \dfrac{s-5}{(s + 6)^2 + 36}\\
\amp = \dfrac{s-5}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{(s+6-6)-5}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{(s+6)-6-5}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{(s+6)-11}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{s+6}{(s + 6)^2 + 6^2} + \dfrac{-11}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{s+6}{(s + 6)^2 + 6^2} -11\cdot \dfrac{1}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{s+6}{(s + 6)^2 + 6^2} - \dfrac{6}{6} \cdot 11\cdot \dfrac{1}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{s+6}{(s + 6)^2 + 6^2} - \dfrac{1}{6} \cdot 11\cdot \dfrac{6}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{s+6}{(s + 6)^2 + 6^2} - \dfrac{11}{6}\cdot \dfrac{6}{(s + 6)^2 + 6^2}\\
\amp = \dfrac{s - a}{(s - a)^2 +b^2} - \dfrac{11}{6}\cdot \dfrac{b}{(s - a)^2 +b^2},
\end{align*}
✍🏻 Problems
12.5.1.
Solution.
We begin by taking the Laplace transform and substituting in the initial conditions.
\begin{align*}
\Big[ s^2X(s) - sx(0) - x'(0) \Big] +6\Big[ sX(s) - x(0) \Big] + 9X(s) \amp = \dfrac{64}{s-5} \mbox{ (by L12,\DL1A and L2)} \\
\Big[ s^2X(s) - s\cdot 3 - 0 \Big] +6\Big[ sX(s) - 3 \Big] + 9X(s) \amp = \dfrac{64}{s-5} \\
s^2X(s) - 3s + 6sX(s) - 18 + 9X(s) \amp = \dfrac{64}{s-5} \\
s^2X(s) + 6sX(s)+9X(s) -3s - 18 \amp = \dfrac{64}{s-5} \text{.}
\end{align*}
Now we aim to solve for \(X(s)\text{,}\) so we use algebra to rearrange as follows:
\begin{align*}
s^2X(s) + 6sX(s)+9X(s) \amp = \dfrac{64}{s-5} + 3s + 18 \\
X(s) \amp = \dfrac{64}{s-5} + (3s + 18)\cdot \dfrac{s-5}{s-5} \\
\amp = \dfrac{64}{s-5} + \dfrac{(3s+18)(s-5)}{s-5} \\
\amp = \dfrac{64}{s-5} + \dfrac{3s^2 +3s- 90}{s-5} \\
\amp = \dfrac{64 + 3s^2 +3s- 90}{s-5} \\
\amp = \dfrac{3s^2 +3s- 26}{s-5} \\
X(s) \amp = \dfrac{3s^2 +3s- 26}{(s-5)(s^2 + 6s + 9)} \\
\amp = \dfrac{3s^2 +3s- 26}{(s-5)(s+3)^2} \text{.}
\end{align*}
We need to do a partial fraction decomposition before we can take the inverse LT.
\begin{align*}
\dfrac{3s^2 +3s- 26}{(s-5)(s+3)^2} \amp = \dfrac{A}{s-5} + \dfrac{B}{s+3} + \dfrac{C}{(s+3)^2} \\
3s^2 +3s - 26 \amp = A(s+3)^2 + B(s-5)(s+3) + C(s-5) \\
\ \amp = A(s^2 + 6s + 9) + B(s^2 - 2s - 15) + Cs - 5C \\
\amp = As^2 + 6As + 9A + Bs^2 - 2Bs - 15B + Cs - 5C \\
\amp = (A + B)s^2 + (6A - 2B + C)s + (9A - 15B - 5C) \text{.}
\end{align*}
Equating the coefficients of like-terms yields, we can solve for \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)
\begin{align*}
A+B \amp = 3 \amp 6A - 2B + C \amp = 3 \amp 9A - 15B - 5C \amp = -26 \\
B \amp = 3-A \amp \amp \amp \amp \\
\amp \amp6A - 2(3-A)+C \amp = 3 \amp \amp \\
\amp \amp 6A-6+2A+C \amp = 3 \amp \amp \\
\amp \amp 8A + C \amp = 9 \amp \amp \\
\amp \amp C \amp = 9 - 8A \amp \amp \\
\amp \amp \amp \amp 9A - 15(3-A) - 5(9-8A) \amp = -26 \\
\amp \amp \amp \amp 9A - 45 + 15A - 45 + 40A \amp = -26 \\
\amp \amp \amp \amp 64A \amp = 64 \\
\amp \amp \amp \amp A \amp = 1 \\
\amp \amp C \amp = 9-8(1) \amp \amp \\
\amp \amp \amp = 1 \amp \amp \\
B \amp = 3 - 1 \amp \amp \amp \amp \\
\amp = 2 \amp \amp \amp \amp \text{.}
\end{align*}
Hence we have \(X(s) = \dfrac{1}{s-5} + \dfrac{2}{s+3} + \dfrac{1}{(s+3)^2}.\) Now we need only take the inverse LT to find the solution.
\begin{align*}
x(t) \amp = \lap^{-1}\left\{ \dfrac{1}{s-5} + \dfrac{2}{s+3} + \dfrac{1}{(s+3)^2} \right\} \\
\amp = \lap^{-1}\left\{ \dfrac{1}{s-5} \right\} + 2\cdot \lap^{-1}\left\{\dfrac{1}{s+3} \right\} + \lap^{-1}\left\{\dfrac{1}{(s+3)^2} \right\} \\
\amp = e^{5t} + 2e^{-3t} + te^{-3t} \text{.}
\end{align*}
Now we aim to solve for \(X(s)\text{,}\) so we use algebra to rearrange as follows.
\begin{align*}
s^2X(s) + 6sX(s)+9X(s) \amp = \dfrac{64}{s-5} + 3s + 18 \\
X(s) \amp = \dfrac{64}{s-5} + (3s + 18)\cdot \dfrac{s-5}{s-5} \\
\amp = \dfrac{64}{s-5} + \dfrac{(3s+18)(s-5)}{s-5} \\
\amp = \dfrac{64}{s-5} + \dfrac{3s^2 +3s- 90}{s-5} \\
\amp = \dfrac{64 + 3s^2 +3s- 90}{s-5} \\
\amp = \dfrac{3s^2 +3s- 26}{s-5} \\
X(s) \amp = \dfrac{3s^2 +3s- 26}{(s-5)(s^2 + 6s + 9)} \\
\amp = \dfrac{3s^2 +3s- 26}{(s-5)(s+3)^2}
\end{align*}
We need to do a partial fraction decomposition before we can take the inverse LT.
\begin{align*}
\dfrac{3s^2 +3s- 26}{(s-5)(s+3)^2} \amp = \dfrac{A}{s-5} + \dfrac{B}{s+3} + \dfrac{C}{(s+3)^2} \\
3s^2 +3s - 26 \amp = A(s+3)^2 + B(s-5)(s+3) + C(s-5) \\
\ \amp = A(s^2 + 6s + 9) + B(s^2 - 2s - 15) + Cs - 5C \\
\amp = As^2 + 6As + 9A + Bs^2 - 2Bs - 15B + Cs - 5C \\
\amp = (A + B)s^2 + (6A - 2B + C)s + (9A - 15B - 5C)
\end{align*}
Equating the coefficients of like terms yields that we can solve for \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)
\begin{align*}
A+B \amp = 3 \amp 6A - 2B + C \amp = 3 \amp 9A - 15B - 5C \amp = -26 \\
B \amp = 3-A \amp \amp \amp \amp \\
\amp \amp6A - 2(3-A)+C \amp = 3 \amp \amp \\
\amp \amp 6A-6+2A+C \amp = 3 \amp \amp \\
\amp \amp 8A + C \amp = 9 \amp \amp \\
\amp \amp C \amp = 9 - 8A \amp \amp \\
\amp \amp \amp \amp 9A - 15(3-A) - 5(9-8A) \amp = -26 \\
\amp \amp \amp \amp 9A - 45 + 15A - 45 + 40A \amp = -26 \\
\amp \amp \amp \amp 64A \amp = 64 \\
\amp \amp \amp \amp A \amp = 1 \\
\amp \amp C \amp = 9-8(1) \amp \amp \\
\amp \amp \amp = 1 \amp \amp \\
B \amp = 3 - 1 \amp \amp \amp \amp \\
\amp = 2 \amp \amp \amp \amp
\end{align*}
Hence we have \(X(s) = \dfrac{1}{s-5} + \dfrac{2}{s+3} + \dfrac{1}{(s+3)^2}. \) Now we need only take the inverse LT to find the solution.
\begin{align*}
x(t) \amp = \lap^{-1}\left\{ \dfrac{1}{s-5} + \dfrac{2}{s+3} + \dfrac{1}{(s+3)^2} \right\} \\
\amp = \lap^{-1}\left\{ \dfrac{1}{s-5} \right\} + 2\cdot \lap^{-1}\left\{\dfrac{1}{s+3} \right\} + \lap^{-1}\left\{\dfrac{1}{(s+3)^2} \right\} \\
\amp = e^{5t} + 2e^{-3t} + te^{-3t}
\end{align*}
12.5.2.
Solution.
We begin by taking the Laplace transform and substituting in the initial conditions.
\begin{align*}
\Big[ s^2Y(s) - s\cdot 2 - 12) \Big] - 2\Big[ sY(s) - 2 \Big] + 5Y(s) \amp = \dfrac{-8}{s+1} \\
s^2Y(s) - 2s - 12 - 2sY(s) +4 + 5Y(s) \amp = \dfrac{-8}{s+1} \\
s^2Y(s) - 2sY(s) + 5Y(s)- 2s - 8 \amp = \dfrac{-8}{s+1} \text{.}
\end{align*}
Now we aim to solve for\(Y(s)\text{,}\) so we use algebra to rearrange as follows.
\begin{align*}
s^2Y(s) - 2sY(s) + 5Y(s) \amp = \dfrac{-8}{s+1} + 2s - 8 \\
(s^2 - 2s + 5)Y(s) \amp = \dfrac{-8}{s+1} + (2s - 8)\cdot \dfrac{s+1}{s+1} \\
(s^2 - 2s + 5)Y(s) \amp = \dfrac{-8}{s+1} + \dfrac{2s^2+10s+8}{s+1} \\
(s^2 - 2s + 5)Y(s) \amp = \dfrac{-8 + 2s^2+10s+8}{s+1} \\
(s^2 - 2s + 5)Y(s) \amp = \dfrac{2s^2+10s}{s+1} \\
Y(s) \amp = \dfrac{2s^2+10s}{(s+1)(s^2 - 2s + 5)} \text{.}
\end{align*}
We need to do a partial fraction decomposition before we can take the inverse LT.
\begin{align*}
\dfrac{2s^2+10s}{(s+1)(s^2 - 2s + 5)} \amp = \dfrac{A}{s+1} + \dfrac{Bs+C}{s^2 - 2s + 5} \\
2s^2 + 10s \amp = A(s^2 - 2s + 5) + (Bs+ C)(s+1) \\
\amp = As^2 - 2As + 5A + Bs^2 + Bs + Cs + C \\
\amp = (A + B)s^2 + (-2A + B + C)s + (5A + C)
\end{align*}
Equating the coefficients of like terms lets us solve for \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)
\begin{align*}
A+B \amp = 2 \amp -2A+B+C \amp = 10 \amp 5A + C \amp = 0 \\
\amp\amp \amp \amp C \amp = -5A \\
\amp\amp \amp \amp \amp \amp -2A + B + (-5A) \amp = 10 \amp\amp \\
\amp\amp -7A + B\amp = 10 \amp \amp \\
\amp\amp B \amp = 10+7A \amp \amp \\
\amp\amp \amp \amp A + (10+7A) \amp = 2 \amp\amp\amp\amp \\
8A+10\amp = 2 \amp \amp \amp \amp \\
8A \amp = -8 \amp \amp \amp \amp \\
A \amp = -1 \amp \amp \amp \amp \\
\amp\amp B \amp = 3 \amp \amp \\
\amp\amp \amp \amp C \amp = 5
\end{align*}
Hence, we have
\begin{equation}
Y(s) = \dfrac{-1}{s+1} + \dfrac{3s+5}{s^2 - 2s + 5}.\tag{12.3}
\end{equation}
We need to take the inverse Laplace transform to find \(y(t)\text{.}\) We need to complete the square on the irreducible quadratic term as follows.
\begin{align*}
s^2 - 2s + 5 \amp = (s^2 - 2s) + 5 \\
\amp = (s^2 - 2s + 1) - 1 + 5 \\
\amp = (s - 1)^2 + 4
\end{align*}
We need to transform the second term, as follows, so we can use L8 and L7 on the Laplace transform table.
\begin{align*}
\dfrac{3s+5}{s^2 - 2s + 5} \amp = \dfrac{3s+5}{(s - 1)^2 + 2^2} \\
\amp = \dfrac{3(s-1)+3+5}{(s - 1)^2 + 2^2} \\
\amp = \dfrac{3(s-1)+8}{(s - 1)^2 + 2^2} \\
\amp = \dfrac{3(s-1)}{(s - 1)^2 + 2^2} + \dfrac{8}{(s - 1)^2 + 2^2} \\
\amp = 3\cdot\dfrac{s-1}{(s - 1)^2 + 2^2} + \dfrac{4\cdot 2}{(s - 1)^2 + 2^2} \\
\amp = 3\cdot\dfrac{s-1}{(s - 1)^2 + 2^2} + 4\cdot\dfrac{2}{(s - 1)^2 + 2^2}
\end{align*}
We are now prepared to take the inverse LT of (12.3) to get the solution to the IVP.
\begin{align*}
y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\
\amp = \lap^{-1}\left\{ \dfrac{-1}{s+1} + \dfrac{3s+5}{s^2 - 2s + 5} \right\} \\
\amp = \lap^{-1}\left\{ -1\cdot \dfrac{1}{s+1} + 3\cdot\dfrac{s-1}{(s - 1)^2 + 2^2} + 4\cdot\dfrac{2}{(s - 1)^2 + 2^2} \right\} \\
\amp = -1\cdot \lap^{-1}\left\{ \dfrac{1}{s+1}\right\} + 3\cdot\lap^{-1}\left\{\dfrac{s-1}{(s - 1)^2 + 2^2} \right\} + 4\cdot\lap^{-1}\left\{\dfrac{2}{(s - 1)^2 + 2^2} \right\} \\
\amp = -e^{-t} + 3e^t\cos(2t) + 4e^t\sin(2t)
\end{align*}
12.5.3.
Solution.
Apply the Laplace transform to both sides of the equation:
\begin{equation*}
\lap{y'' + 3y' + 2y} = \lap{0}\text{.}
\end{equation*}
Using the properties of the Laplace transform for derivatives, we have:
\begin{equation*}
s^2Y(s) - sy(0) - y'(0) + 3[sY(s) - y(0)] + 2Y(s) = 0\text{.}
\end{equation*}
Substitute the initial conditions \(y(0) = 2\) and \(y'(0) = -1\text{:}\)
\begin{equation*}
s^2Y(s) - 2s + 1 + 3sY(s) - 6 + 2Y(s) = 0\text{.}
\end{equation*}
Solve for \(Y(s)\text{:}\)
\begin{equation*}
Y(s) = \dfrac{2s + 5}{s^2 + 3s + 2}\text{.}
\end{equation*}
Factor the denominator:
\begin{equation*}
Y(s) = \dfrac{2s + 5}{(s + 1)(s + 2)}\text{.}
\end{equation*}
Use partial fraction decomposition to find the inverse Laplace transform:
\begin{equation*}
Y(s) = \dfrac{A}{s + 1} + \dfrac{B}{s + 2}\text{.}
\end{equation*}
Solve for \(A\) and \(B\text{:}\)
\begin{equation*}
Y(s) = \dfrac{1}{s + 1} - \dfrac{3}{s + 2}\text{.}
\end{equation*}
Finally, take the inverse Laplace transform to find \(y(t)\text{:}\)
\begin{equation*}
y(t) = e^{-t} - 3e^{-2t}\text{.}
\end{equation*}
12.5.4.
Solution.
Apply the Laplace transform to both sides:
\begin{equation*}
\lap{y' + 4y} = \lap{10e^{-2t}}\text{.}
\end{equation*}
Use the properties of the Laplace transform:
\begin{equation*}
sY(s) - y(0) + 4Y(s) = \dfrac{10}{s + 2}\text{.}
\end{equation*}
Substitute the initial condition \(y(0) = 3\text{:}\)
\begin{equation*}
(s + 4)Y(s) - 3 = \dfrac{10}{s + 2}\text{.}
\end{equation*}
Solve for \(Y(s)\text{:}\)
\begin{equation*}
Y(s) = \dfrac{10}{(s + 4)(s + 2)} + \dfrac{3}{s + 4}\text{.}
\end{equation*}
Use partial fraction decomposition:
\begin{equation*}
Y(s) = \dfrac{A}{s + 4} + \dfrac{B}{s + 2}\text{.}
\end{equation*}
Find \(A\) and \(B\text{:}\)
\begin{equation*}
A = \dfrac{7}{2}, B = -\dfrac{5}{2}\text{.}
\end{equation*}
The inverse Laplace transform gives:
\begin{equation*}
y(t) = \dfrac{7}{2}e^{-4t} - \dfrac{5}{2}e^{-2t}\text{.}
\end{equation*}
12.5.5.
Solution.
Applying the Laplace transform to both sides, we get:
\begin{equation*}
sY(s) - y(0) + 3Y(s) = \dfrac{6}{s-2}
\end{equation*}
Substituting the initial condition \(y(0) = 1\text{,}\) the equation becomes:
\begin{equation*}
sY(s) - 1 + 3Y(s) = \dfrac{6}{s-2}
\end{equation*}
Rearranging and solving for \(Y(s)\text{:}\)
\begin{align*}
Y(s)(s + 3) \amp = 1 + \dfrac{6}{s-2}\\
Y(s) \amp = \dfrac{1}{s+3} + \dfrac{6}{(s-2)(s+3)}
\end{align*}
We can now decompose the second term using partial fractions:
\begin{equation*}
\dfrac{6}{(s-2)(s+3)} = \dfrac{A}{s-2} + \dfrac{B}{s+3}
\end{equation*}
Solving for \(A\) and \(B\text{,}\) we get \(A = 2\) and \(B = 4\text{.}\) Therefore:
\begin{gather*}
Y(s) = \dfrac{1}{s+3} + \dfrac{2}{s-2} + \dfrac{4}{s+3}\\
Y(s) = \dfrac{5}{s+3} + \dfrac{2}{s-2}
\end{gather*}
Taking the inverse Laplace transform, we obtain the solution:
\begin{equation*}
y(t) = 5e^{-3t} + 2e^{2t}
\end{equation*}
12.5.6.
Solution.
Applying the Laplace transform:
\begin{equation*}
s^2Y(s) - sy(0) - y'(0) + 4Y(s) = \dfrac{s}{s^2 + 4}
\end{equation*}
Substituting the initial conditions:
\begin{equation*}
s^2Y(s) - 1 + 4Y(s) = \dfrac{s}{s^2 + 4}
\end{equation*}
Solving for \(Y(s)\text{:}\)
\begin{equation*}
Y(s)(s^2 + 4) = 1 + \dfrac{s}{s^2 + 4}
\end{equation*}
After rearranging and solving, the inverse Laplace transform gives:
\begin{equation*}
y(t) = \dfrac{t+2}{4}\sin(2t)
\end{equation*}
12.5.7.
12.5.8.
12.5.9.
Solution.
We intend to use Laplace transforms to solve this IVP, but we need to verify that this is an appropriate technique. We can verify that this is
-
The DE is linear,
-
The DE has constant coefficients, and
-
Initial conditions are provided.
Hence, it is appropriate to proceed with the Laplace transform solution technique. We also note that \(y\) is the dependent variable and \(t\) is the independent variable, so our goal is to find a function \(y(t)\) that satisfies the DE and the initial conditions.
📝
We begin by taking the Laplace transform of both sides of the DE, using linearity as needed.
\begin{align*}
\lap{y'' + 3y' + 2y} \amp = \lap{ 4t }\\
\lap{y''} + 3\lap{y'} + 2\lap{y} \amp = 4\cdot\lap{ t }\\
\Big[s^2Y(s) - sy(0) - y'(0) \Big] + 3\Big[ sY(s) - y(0) \Big] + 2Y(s) \amp = \dfrac{4}{s^2}
\end{align*}
Now we use the provided initial conditions.
\begin{align*}
\Big[s^2Y(s) - s\cdot 1 - 0 \Big] + 3\Big[ sY(s) - 1 \Big] + 2Y(s) \amp = \dfrac{4}{s^2}\\
s^2Y(s) - s + 3sY(s) - 3 + 2Y(s) \amp = \dfrac{4}{s^2}
\end{align*}
Next we will use algebra to solve for \(Y(s).\)
\begin{align*}
s^2Y(s)+ 3sY(s) + 2Y(s) \amp = \dfrac{4}{s^2} + s + 3\\
Y(s)\Big[ s^2 + 3s + 2 \Big] \amp = \dfrac{4}{s^2} + \dfrac{s^3}{s^2} + \dfrac{3s^2}{s^2}\\
\amp = \dfrac{4 + s^3 + 3s^2}{s^2}\\
Y(s) \amp = \dfrac{s^3 + 3s^2 + 4}{s^2(s^2 + 3s + 2)}\\
\amp = \dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)}
\end{align*}
We need to find the inverse Laplace transform of both sides of the equation. In order to do that, we apply partial fraction decomposition to the rational function on the right hand side, giving
\begin{equation*}
\dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} = \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1}.
\end{equation*}
We have a repeated linear factor, \(s^2\text{,}\) and two other linear factors, so the form of the partial fraction decomposition can be written as
\begin{equation*}
\dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} = \dfrac{A}{s} + \dfrac{B}{s^2} + \dfrac{C}{s+2} + \dfrac{D}{s+1}
\end{equation*}
\begin{align*}
s^3 + 3s^2 + 4 \amp = A(s)(s+2)(s+1) + B(s+2)(s+1) + C(s^2)(s+1) + D(s^2)(s+2)\\
\amp = As(s^2 + 3s + 2) + B(s^2 + 3s + 2) + Cs^3 + Cs^2 + Ds^3 + 2Ds^2\\
\amp = As^3 + 3As^2 + 2As + Bs^2 + 3Bs + 2B + Cs^3 + Cs^2 + Ds^3 + 2Ds^2\\
s^3 + 3s^2 + 0s + 4
\amp = (A + C + D)s^3 + (3A + B + C+ 2D)s^2 + (2A + 3B)s + (2B)
\end{align*}
\begin{align*}
A+C+D \amp = 1 \amp 3A+B+C+2D \amp = 3 \amp 2A+3B \amp = 0 \amp 2B \amp = 4 \\
\amp \amp \amp \amp \amp \amp B \amp = 2 \\
\amp \amp 3A+2+C+2D \amp = 3 \amp 2A+ 3(2)\amp = 0 \amp \amp \\
\amp \amp 3A+C+2D \amp = 1 \amp 2A \amp = -6 \amp \amp \\
\amp \amp \amp \amp A \amp = -3 \amp \amp \\
-3+C+D \amp = 1 \amp 3(-3)+C+ 2D \amp = 1 \amp \amp \amp \amp \\
C+D \amp = 4 \amp C+2D \amp = 10 \amp \amp \amp \amp \\
C \amp = 4 - D\amp \amp \amp \amp \amp \amp \\
\amp \amp (4-D) + 2D \amp = 10 \amp \amp \amp \amp \\
\amp \amp 4 + D \amp = 10 \amp \amp \amp \amp \\
\amp \amp D \amp = 6 \amp \amp \amp \amp \\
C \amp = 4-6 \amp \amp \amp \amp \amp \amp \\
\amp = -2 \amp \amp \amp \amp \amp \amp
\end{align*}
Hence, we have
\begin{equation*}
\dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} = \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1}
\end{equation*}
We are now prepared to take the inverse Laplace transform.
\begin{align*}
Y(s) \amp = \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1} \\
\ilap{Y(s)} \amp = \ilap{ \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1} } \\
y(t) \amp = -3\ilap{ \dfrac{1}{s} } + 2\ilap{ \dfrac{1}{s^2} } - 2\ilap{ \dfrac{1}{s+2} } + 6\ilap{ \dfrac{1}{s+1} }\\
\amp = -3 + 2t - 2e^{-2t} + 6e^{-t}
\end{align*}
Thus, we have found the desired unknown function
\begin{equation*}
y(t) = -3 + 2t - 2e^{-2t} + 6e^{-t}.
\end{equation*}
Optional: verify the solution
\begin{align*}
y'(t) \amp = 0 + 2 + 4e^{-2t} - 6e^{-t}\\
y''(t) \amp = 0 - 8e^{-2t}+ 6e^{-t}
\end{align*}
\begin{align*}
LHS \amp = y'' + 3y' + 2y\\
\amp = \left( - 8e^{-2t}+ 6e^{-t} \right) \\
\amp = -8e^{-2t}+ 6e^{-t}\\
\amp = 4t\\
\amp = RHS
\end{align*}
We also verify the initial conditions:
\begin{align*}
y(0) \amp = -3 + 2(0) - 2e^{-2(0)} + 6e^{-0}\\
\amp = -3 + 0 - 2+ 6\\
\amp = 1\\
\amp\\
y'(0) \amp = 2 + 4e^{-2(0)} - 6e^{-0}\\
\amp = 2 + 4 - 6\\
\amp = 0
\end{align*}
12.5.10.
Solution.
Step 1— Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{gather*}
\lap{y'' + 3y' + 2y} = 0\\
\left(s^2Y(s) - s\right) + 3\left(sY(s) - 1\right) + 2Y(s) = 0
\end{gather*}
Step 2— Solve & Prepare \(Y(s)\). Isolating \(Y(s)\text{:}\)
\begin{gather*}
\left(s^2 + 3s + 2\right)Y(s) = s + 3\\
Y(s) = \dfrac{s + 3}{(s + 1)(s + 2)} = \dfrac{2}{s + 1} - \dfrac{1}{s + 2}
\end{gather*}
Step 3— Inverse Transform. Applying the inverse Laplace transform:
\begin{equation*}
y(t) = 2e^{-t} - e^{-2t}.
\end{equation*}
12.5.11.
Solution.
Step 1— Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{gather*}
\lap{y'' - y} = \lap{e^{2t}}\\
\left(s^2Y(s) - 1\right) - Y(s) = \dfrac{1}{s - 2}
\end{gather*}
Step 2— Solve & Prepare \(Y(s)\). Isolating \(Y(s)\text{:}\)
\begin{gather*}
\left(s^2 - 1\right)Y(s) = 1 + \dfrac{1}{s - 2} = \dfrac{s - 1}{s - 2}\\
Y(s) = \dfrac{1}{(s - 2)(s + 1)} = \dfrac{1}{3}\left(\dfrac{1}{s - 2} - \dfrac{1}{s + 1}\right)
\end{gather*}
Step 3— Inverse Transform. Applying the inverse Laplace transform:
\begin{equation*}
y(t) = \dfrac{1}{3}e^{2t} - \dfrac{1}{3}e^{-t}.
\end{equation*}
12.5.12.
Solution.
Step 1— Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{gather*}
\lap{y' + y} = \lap{4}\\
\lap{y'} + \lap{y} = \lap{4}\\
sY(s) - 2 + Y(s) = \dfrac{4}{s}
\end{gather*}
Step 2— Solve & Prepare \(Y(s)\). Next, we isolate \(Y(s)\) on one side of the equation:
\begin{align*}
sY(s) - 2 + Y(s) \amp = \dfrac{4}{s} \\
(s + 1)Y(s) \amp = \dfrac{4}{s} + 2 \\
Y(s) \amp = \dfrac{1}{s + 1}\left(\dfrac{4}{s} + 2\right) \\
Y(s) \amp = \ub{\dfrac{4}{s(s + 1)}}_{(*)} + \ub{\dfrac{2}{s + 1}}_{(**)}
\end{align*}
Step 2— Solve & Prepare \(Y(s)\). We need to express \(Y(s)\) as a sum of functions that match known forms in the Laplace transform table. We see from the last equation, \((**)\) is ready to go, but \((*)\) requires partial fraction decomposition. We start by writing down the form of the decomposition,
\begin{equation*}
(*) = \dfrac{4}{s(s + 1)} = \dfrac{A}{s} + \dfrac{B}{s + 1} \quad \text{or}
\end{equation*}
\begin{equation*}
4 = A(s + 1) + Bs
\end{equation*}
Now, we find \(A\) and \(B\) by plugging in values of \(s\text{:}\)
| \(s=0\) : | \(4\) | \(=\) | \(A(0 + 1) + B(0)\) |
| \(4\) | \(=\) | \(A\) | |
| \(s=-1\) : | \(4\) | \(=\) | \(A(-1 + 1) + B(-1)\) |
| \(-4\) | \(=\) | \(B\) |
Thus, the prepared \(Y(s)\) is
\begin{equation*}
Y(s) = \ub{\dfrac{4}{s} + \dfrac{-4}{s + 1}}_{(*)} + \ub{\dfrac{2}{s + 1}}_{(**)}
\end{equation*}
Step 3— Inverse Transform. Finally, we apply the backward step to get the solution:
\begin{align*}
\ilap{Y(s)} \amp = \ilap{\dfrac{4}{s} + \dfrac{-4}{s + 1}}\\
y(t) \amp = 4 - 4e^{-t}
\end{align*}
12.5.13.
Solution.
Step 1— Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{align*}
\lap{y'' - 3y' + 2y} \amp = \lap{e^{2t}}\\
\lap{y''} - 3\lap{y'} + 2\lap{y} \amp = \dfrac{1}{s - 2}
\end{align*}
where \(\lap{y(t)} = Y(s)\) and
-
\(\lap{y'} = sY(s) - y(0) = sY(s) - 1\text{,}\)
-
\(\lap{y''} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - s\text{.}\)
So the complete forward transform is
\begin{equation*}
s^2Y(s) - s - 3[sY(s) - 1] + 2Y(s) = \dfrac{1}{s - 2}
\end{equation*}
Step 2— Solve & Prepare \(Y(s)\). Next, we isolate \(Y(s)\) on one side of the equation:
\begin{gather*}
s^2Y(s) - s - 3sY(s) + 3 + 2Y(s) = \dfrac{1}{s - 2}\\
[s^2 - 3s + 2]Y(s) - s + 3 = \dfrac{1}{s - 2}\\
Y(s)[(s - 1)(s - 2)] = \dfrac{1}{s - 2} + s - 3\\
Y(s) = \dfrac{1}{(s - 1)(s - 2)}\ub{\left(\dfrac{1}{s - 2} + s - 3\right)}_{(*)}
\end{gather*}
Step 2— Solve & Prepare \(Y(s)\). Now, we need to express \(Y(s)\) as a sum of functions that match known forms in the Laplace transform table. This can be simplified slightly by first combining the terms in \((*)\) as a single fraction, like so
\begin{equation*}
(*) = \dfrac{1}{s - 2} + \dfrac{(s - 3)(s - 2)}{s - 2}
= \dfrac{1 + s^2 - 5s + 6}{s - 2} = \dfrac{s^2 - 5s + 7}{s - 2}\text{.}
\end{equation*}
Plugging this back into the equation for \((*)\) gives the new \(Y(s)\text{,}\)
\begin{equation*}
Y(s) = \dfrac{1}{(s - 1)(s - 2)}\cdot \dfrac{s^2 - 5s + 7}{s - 2} = \dfrac{s^2 - 5s + 7}{(s - 1)(s - 2)^2}
\end{equation*}
and we are ready to apply partial fraction decomposition. The form of the decomposition is
\begin{align*}
\dfrac{s^2 - 5s + 7}{(s - 1)(s - 2)^2}
\amp = \dfrac{A}{s - 1} + \dfrac{B}{s - 2} + \dfrac{C}{(s - 2)^2} \quad \text{or}\\
s^2 - 5s + 7
\amp = A(s - 2)^2 + B(s - 1)(s - 2) + C(s - 1).
\end{align*}
Now, we find \(A, B, \) and \(C\) by plugging in values of \(s\text{,}\)
| \(s=1\) : | \(1 - 5 + 7\) | \(=\) | \(A(1 - 2)^2 + B(1 - 1)(1 - 2) + C(1 - 1)\) |
| \(3\) | \(=\) | \(A\) | |
| \(s=2\) : | \(4 - 10 + 7\) | \(=\) | \(3(2 - 2)^2 + B(2 - 1)(2 - 2) + C(2 - 1)\) |
| \(1\) | \(=\) | \(C\) | |
| \(s=0\) : | \(7\) | \(=\) | \(3(0 - 2)^2 + B(0 - 1)(0 - 2) + 1(0 - 1)\) |
| \(7\) | \(=\) | \(12 + 2B - 1\) | |
| \(-2\) | \(=\) | \(B\) |
Thus, the prepared \(Y(s)\) is
\begin{equation*}
Y(s) = \dfrac{3}{s - 1} + \dfrac{-2}{s - 2} + \dfrac{1}{(s - 2)^2}
\end{equation*}
Step 3— Inverse Transform. Finally, we perform the backward step to get the solution:
\begin{align*}
\ilap{Y(s)}\\
\amp = \ilap{\dfrac{3}{s - 1}} + \ilap{\dfrac{-2}{s - 2}} + \ilap{\dfrac{1}{(s - 2)^2}}\\
y(t) \amp = 3\, e^{t} - 2\, e^{2t} + t\, e^{2t}
\end{align*}
12.5.14.
Solution.
Step 1— Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{gather*}
\lap{x'' - 4x' + 13x} = \lap{54e^{-t}}\\
\lap{x''} - 4\lap{x'} + 13\lap{x} = \dfrac{54}{s + 1}
\end{gather*}
where \(\lap{x(t)} = X(s)\) and
-
\(\lap{x'} = sX(s) - x(0) = sX(s)\text{,}\)
-
\(\lap{x''} = s^2X(s) - sx(0) - x'(0) = s^2X(s)\text{.}\)
So the complete forward transform is:
\begin{equation*}
s^2X(s) - 4sX(s) + 13X(s) = \dfrac{54}{s + 1}
\end{equation*}
Step 2— Solve & Prepare \(Y(s)\). Next, we isolate \(X(s)\) on one side of the equation:
\begin{gather*}
s^2X(s) - 4sX(s) + 13X(s) = \dfrac{54}{s + 1}\\
[s^2 - 4s + 13]X(s) = \dfrac{54}{s + 1}\\
X(s) = \dfrac{54}{(s + 1)(s^2 - 4s + 13)}
\end{gather*}
Step 2— Solve & Prepare \(Y(s)\). Now, we need to express \(X(s)\) as a sum of functions that match known forms in the Laplace transform table, which requires partial fraction decomposition. We start by writing down the form of the decomposition,
\begin{gather*}
\dfrac{54}{(s + 1)(s^2 - 4s + 13)} = \dfrac{A}{s + 1} + \dfrac{Bs + C}{s^2 - 4s + 13} \quad \text{or}\\
54 = A(s^2 - 4s + 13) + (Bs + C)(s + 1)
\end{gather*}
Now, we find \(A, B, \) and \(C\) by plugging in values of \(s\text{,}\)
So the updated \(X(s)\) is
\begin{equation*}
X(s) = \dfrac{37}{s + 1} + \dfrac{585s - 427}{s^2 - 4s + 13}\text{.}
\end{equation*}
Note, the second term is not yet ready, and we need to complete the square of its denominator before we can do the backward step.
\begin{align*}
X(s) \amp = \dfrac{37}{s + 1} + \dfrac{585s - 427}{(s - 2)^2 + 9}\\
\amp = \dfrac{37}{s + 1} + \dfrac{585s - 2(585) + 2(585)- 427}{(s - 2)^2 + 9}\\
\amp = \dfrac{37}{s + 1} + \dfrac{585s - 2(585)}{(s - 2)^2 + 9} + \dfrac{2(585)- 427}{(s - 2)^2 + 9}\\
\amp = \dfrac{37}{s + 1} + 585\dfrac{s - 2}{(s - 2)^2 + 9} + 743\dfrac{1}{(s - 2)^2 + 9}\\
\amp = \dfrac{37}{s + 1} + 585\dfrac{s - 2}{(s - 2)^2 + 9} + \dfrac{743}{3}\dfrac{3}{(s - 2)^2 + 9}
\end{align*}
Step 3— Inverse Transform. Now for the backward step to get the solution:
\begin{align*}
x(t) \amp = 37e^{-t} + 585\cos(3t) + \dfrac{743}{3}\sin(3t)
\end{align*}
13 Piecewise Forcing Functions
13.7 Exercises
🏋️♂️ Practice Drills
✍🏻 Problems
13.7.1.
Solution.
This is already in the form \(e^{-cs}F(s)\) with \(c = 3\) and \(F(s) = \dfrac{2}{s^2 + 9}\text{.}\)
The inverse transform of \(\dfrac{2}{s^2 + 9}\) is \(\sin(3t)\text{,}\) since this matches the standard form:
\begin{equation*}
\ilap{ \dfrac{b}{s^2 + b^2} } = \sin(bt)\text{.}
\end{equation*}
Applying the shift rule:
\begin{equation*}
y(t) = u_3(t) \cdot \sin(3(t - 3)) = u_3(t)\cdot \sin(3t - 9)
\end{equation*}
13.7.2.
Solution.
We identify this as \(e^{-2s} \cdot F(s)\text{,}\) where
\begin{equation*}
F(s) = \dfrac{s}{s^2 + 4}.
\end{equation*}
This matches the form for \(\cos(2t)\text{:}\)
\begin{equation*}
\ilap{ \dfrac{s}{s^2 + b^2} } = \cos(bt).
\end{equation*}
So the inverse is:
\begin{equation*}
f(t) = u_2(t) \cdot \cos(2(t - 2)) = u_2(t)\cdot \cos(2t - 4)
\end{equation*}
13.7.7.
Solution.
🧠 Derivation 271. Roadmap-Summary.
Step 1: Apply the Laplace Transform.
Step 2: Solve for \(Y(s)\).
Step 3: Prepare for the Inverse Transform.
The exponential factor \(e^{-3s}\) indicates a delay. Focus on the base transform:
\begin{equation*}
\dfrac{1}{s(s^2 + 4)}
\end{equation*}
From the table, this matches the transform of \(1 - \cos(2t)\text{.}\) Therefore:
\begin{equation*}
\ilap{ \dfrac{1}{s(s^2 + 4)} } = 1 - \cos(2t)
\end{equation*}
The delay is applied using the shift theorem.
Step 4: Take the Inverse Transform.
\begin{gather*}
y(t) = \ilap{ \dfrac{e^{-3s}}{s(s^2 + 4)} } = u_3(t)\left[1 - \cos(2(t - 3))\right]
\end{gather*}
Therefore, the solution is:
\begin{equation*}
y(t) = u_3(t)\left(1 - \cos(2(t - 3))\right)
\end{equation*}
13.7.10.
Solution.
🗺️ Summary 272. Roadmap-Summary.
\begin{equation*}
y(t) = \left(\dfrac{1}{3} - \dfrac{1}{6}\cos(3(t-2))\right)u_2(t) - \left(\dfrac{1}{3} - \dfrac{1}{6}\cos(3(t-3))\right)u_3(t)
\end{equation*}
\begin{equation*}
\os{\Large\text{4️⃣}}{\us{\large\text{backward}}{\os{\large\laplacesym^{-1}}{\longleftarrow}}}
\end{equation*}
\begin{equation*}
Y(s) = \dfrac{1}{s(s^2 + 9)}(e^{-2s} - e^{-3s})
\end{equation*}
Prepare the Forcing Function for Step 1.
We begin by rewriting \(h(t)\) using unit step functions. Since \(h(t)\) equals 1 on \([2, 3)\) and 0 elsewhere, we express it as:
\begin{equation*}
h(t) = u_2(t) - u_3(t)
\end{equation*}
This transforms the differential equation into:
\begin{equation*}
y'' + 9y = u_2(t) - u_3(t), \qquad y(0) = 0,\quad y'(0) = 0
\end{equation*}
Step 1: Apply the Laplace Transform to Both Sides.
Apply the Laplace transform to both sides:
\begin{equation*}
\lap{y''} + 9\lap{y} = \lap{u_2(t)} - \lap{u_3(t)}
\end{equation*}
Using standard rules:
\begin{equation*}
s^2Y(s) + 9Y(s) = \dfrac{e^{-2s}}{s} - \dfrac{e^{-3s}}{s}
\end{equation*}
Step 2: Solve for \(Y(s)\).
Factor out \(Y(s)\text{:}\)
\begin{gather*}
(s^2 + 9)Y(s) = \dfrac{e^{-2s}}{s} - \dfrac{e^{-3s}}{s}
\end{gather*}
Then divide:
\begin{equation*}
Y(s) = \dfrac{1}{s^2 + 9} \left( \dfrac{e^{-2s}}{s} - \dfrac{e^{-3s}}{s} \right)
\end{equation*}
Step 3: Prepare \(Y(s)\) for an Inverse Transform.
Step 3 focuses on rewriting \(Y(s)\) to match inverse Laplace rules. We separate the exponentials:
\begin{equation*}
Y(s) = \dfrac{1}{s(s^2 + 9)}\left(e^{-2s} - e^{-3s}\right)
\end{equation*}
Let the rational part be:
\begin{equation*}
F(s) = \dfrac{1}{s(s^2 + 9)}
\end{equation*}
Then:
\begin{equation*}
Y(s) = F(s)\left(e^{-2s} - e^{-3s}\right)
\end{equation*}
This means the inverse will take the form:
\begin{equation*}
y(t) = \ilap{F(s)e^{-2s}} - \ilap{F(s)e^{-3s}}
\end{equation*}
but we must first identify \(f(t) = \ilap{F(s)}\) using partial fractions. The details are shown below:
Solution. Partial Fraction Decomposition of \(F(s)\)
Write:
\begin{equation*}
F(s) = \dfrac{1}{s(s^2 + 9)} = \dfrac{A}{s} + \dfrac{Bs + C}{s^2 + 9}
\end{equation*}
Multiply by \(s(s^2 + 9)\) and group:
\begin{align*}
1 \amp = A(s^2 + 9) + (Bs + C)s\\
\amp = (A + B)s^2 + Cs + 9A
\end{align*}
Match coefficients:
So:
\begin{equation*}
F(s) = \dfrac{1}{9}\left(\dfrac{1}{s} - \dfrac{s}{s^2 + 9}\right)
\end{equation*}
This completes Step 3. We’ll now apply the inverse transform.
Step 4: Invert \(Y(s)\) to Recover \(y(t)\).
We now compute:
\begin{equation*}
y(t) = \ilap{F(s)e^{-2s}} - \ilap{F(s)e^{-3s}}
\end{equation*}
By rule \(L15\), this gives:
\begin{equation*}
y(t) = f(t - 2)\cdot u_2(t) - f(t - 3)\cdot u_3(t)
\end{equation*}
where:
\begin{gather*}
f(t) = \ilap{F(s)} = \dfrac{1}{9}\ilap{\dfrac{1}{s}} - \dfrac{1}{9}\ilap{\dfrac{s}{s^2 + 9}}\\
= \dfrac{1}{9} - \dfrac{1}{27}\cos(3t)
\end{gather*}
Substitute:
\begin{equation*}
y(t) = \left(\dfrac{1}{9} - \dfrac{1}{27}\cos(3(t-2))\right)u_2(t)
- \left(\dfrac{1}{9} - \dfrac{1}{27}\cos(3(t-3))\right)u_3(t)
\end{equation*}
This is the complete solution for all \(t \ge 0\text{.}\)
13.7.11.
Solution.
Step 1 — Into the Laplace Domain.
Recall that our first step is to apply the Laplace transform of both sides:
\begin{equation*}
\lap{y'' + y} = \lap{g(t)}
\end{equation*}
It’s clear how to handle the left-side, but the right-side requires the transform of a piecewise function. Before we can use the rules we derived in this chapter, we first need to write \(g(t)\) in unit step form.
Apply the Laplace transform to both sides of the differential equation:
\begin{equation*}
y'' + y = \left\{
\begin{array}{ll}
0, \amp t \lt 1 \\
3t, \amp 1 \le t \lt 2 \\
0, \amp t \ge 2
\end{array}
\right.,
\qquad
y(0) = 0,\quad y'(0) = 0
\end{equation*}
The forcing function is active only from \(t = 1\) to \(t = 2\text{,}\) so we rewrite it using a unit step window:
\begin{equation*}
g(t) = 3t \cdot \left(u_1(t) - u_2(t)\right).
\end{equation*}
Apply the Laplace transform to both sides of the DE:
\begin{equation*}
\lap{y'' + y} = \lap{3t \cdot (u_1(t) - u_2(t))}.
\end{equation*}
On the left-hand side, using \(y(0) = y'(0) = 0\text{,}\) we have:
\begin{equation*}
s^2 Y(s) + Y(s).
\end{equation*}
On the right-hand side, distribute the Laplace transform:
\begin{equation*}
\lap{3t \cdot (u_1(t) - u_2(t))} = 3\lap{t \cdot u_1(t)} - 3\lap{t \cdot u_2(t)}.
\end{equation*}
Use the shift rule for each term:
\begin{gather*}
\lap{t \cdot u_1(t)} = e^{-s} \lap{t + 1} = e^{-s} \cdot \lap{t + 1} \\
\lap{t \cdot u_2(t)} = e^{-2s} \lap{t + 2} = e^{-2s} \cdot \lap{t + 2}
\end{gather*}
Now apply linearity and known transforms:
\begin{gather*}
\lap{t + c} = \lap{t} + c\lap{1} = \dfrac{1}{s^2} + \dfrac{c}{s}
\end{gather*}
So:
\begin{align*}
\lap{3t \cdot (u_1 - u_2)}
\amp = 3e^{-s}\left(\dfrac{1}{s^2} + \dfrac{1}{s}\right)
- 3e^{-2s}\left(\dfrac{1}{s^2} + \dfrac{2}{s}\right)
\end{align*}
Combining both sides, we arrive at the Laplace equation:
\begin{equation*}
(s^2 + 1)Y(s) = 3e^{-s}\left(\dfrac{1}{s^2} + \dfrac{1}{s}\right)
- 3e^{-2s}\left(\dfrac{1}{s^2} + \dfrac{2}{s}\right).
\end{equation*}
This completes Step 1.
Now that we’ve built the Laplace equation, the remaining steps are familiar. We’ll solve for \(Y(s)\text{,}\) simplify, and invert. But there’s one twist: the presence of \(e^{-cs}\) terms means we’ll need special care when taking the inverse Laplace transform. Let’s explore that next.
Step 2 — Solving in the Laplace Domain.
Step 3 — Leaving the Laplace Domain.
Once we’ve solved for \(Y(s)\) in the Laplace domain, the final challenge is to take the inverse Laplace transform. When the forcing function was continuous, this step relied on partial fractions and matching terms from the Laplace table. That still works here, but now we’ll also see terms involving exponentials like \(e^{-cs}F(s)\text{.}\)
These exponentials arise directly from transforming unit-step expressions. To move back to the \(t\)-domain, we need a new tool: a rule that undoes the exponential shift.
13.7.12.
Solution.
Roadmap-Summary.
\begin{equation*}
y(t) = \left(1 - e^{-2(t - 1)}\right)u_1(t)
- \left(1 - e^{-2(t - 2)}\right)u_2(t)
\end{equation*}
\begin{equation*}
\os{\Large\text{4️⃣}}{\us{\large\text{backward}}{\os{\large\laplacesym^{-1}}{\longleftarrow}}}
\end{equation*}
\begin{equation*}
Y(s) = \dfrac{1}{s(s + 2)}(e^{-s} - e^{-2s})
\end{equation*}
Preparation — Forcing Function in Step Form.
We begin by rewriting \(g(t)\) with unit step functions. Since it equals \(3\) on \([0,1)\text{,}\) \(0\) on \([1,4)\text{,}\) and \(1\) after \(t=4\text{:}\)
\begin{equation*}
g(t) = 3(u_0(t) - u_1(t)) + (1 - 3)(u_4(t))
\end{equation*}
which simplifies to:
\begin{equation*}
g(t) = 3u_0(t) - 3u_1(t) + u_4(t).
\end{equation*}
Step 1 — Into the Laplace Domain.
Take the Laplace transform of both sides:
\begin{equation*}
\lap{y''} + 2\lap{y'} = \lap{3u_0(t) - 3u_1(t) + u_4(t)}.
\end{equation*}
Using the initial conditions \(y(0)=0\) and \(y'(0)=0\text{,}\) the left side becomes:
\begin{equation*}
s^2 Y(s) + 2s Y(s).
\end{equation*}
Apply L\(_9\) to each step term:
\begin{equation*}
\lap{3u_0(t)} = \dfrac{3}{s}, \quad
\lap{-3u_1(t)} = -\dfrac{3e^{-s}}{s}, \quad
\lap{u_4(t)} = \dfrac{e^{-4s}}{s}.
\end{equation*}
This gives the Laplace-domain equation:
\begin{equation*}
(s^2 + 2s)Y(s) = \dfrac{3}{s} - \dfrac{3e^{-s}}{s} + \dfrac{e^{-4s}}{s}.
\end{equation*}
Step 2 — Solve for \(Y(s)\).
\begin{gather*}
Y(s) = \dfrac{1}{s(s + 2)}\left( 3 - 3e^{-s} + e^{-4s} \right).
\end{gather*}
To prepare for inversion, write:
\begin{equation*}
Y(s) = F(s)\big( 3 - 3e^{-s} + e^{-4s} \big),
\end{equation*}
\begin{equation*}
F(s) = \dfrac{1}{s(s + 2)}.
\end{equation*}
Solution. Partial Fraction Decomposition of \(F(s)\)
Decompose:
\begin{equation*}
\dfrac{1}{s(s + 2)} = \dfrac{A}{s} + \dfrac{B}{s + 2}.
\end{equation*}
Multiply through:
\begin{equation*}
1 = A(s + 2) + B(s).
\end{equation*}
Sub in convenient \(s\)-values:
-
At \(s = 0\text{:}\) \(1 = 2A\) → \(A = \dfrac{1}{2}\)
-
At \(s = -2\text{:}\) \(1 = -2B\) → \(B = -\dfrac{1}{2}\)
Thus:
\begin{equation*}
F(s) = \dfrac{1}{2}\dfrac{1}{s} - \dfrac{1}{2}\dfrac{1}{s+2}.
\end{equation*}
Take the inverse Laplace:
\begin{equation*}
f(t) = \ilap{F(s)} = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t}.
\end{equation*}
Step 3 — Back to the Time Domain.
Each exponential factor in \(Y(s)\) corresponds to a shifted version of \(f(t)\) multiplied by the appropriate step function using L\(_11\).
\begin{equation*}
\ilap{Y(s)} = 3f(t) - 3f(t-1)u_1(t) + f(t-4)u_4(t).
\end{equation*}
Substitute \(f(t) = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t}\) into each term:
\begin{equation*}
y(t) = 3\left(\dfrac{1}{2} - \dfrac{1}{2}e^{-2t}\right)
- 3\left(\dfrac{1}{2} - \dfrac{1}{2}e^{-2(t-1)}\right)u_1(t)
+ \left(\dfrac{1}{2} - \dfrac{1}{2}e^{-2(t-4)}\right)u_4(t).
\end{equation*}
This simplifies to the complete solution for all \(t \ge 0\text{.}\) It clearly shows how each piece of the forcing term triggers its own shifted response in the solution.
13.7.13.
Solution.
\begin{align*}
D(s) \amp = \lap{ d(t) } \\
\amp = \int_0^{\infty}e^{-st}d(t) dt \\
\amp = \int_0^3 e^{-st}\cdot 7 dt + \int_3^{\infty} e^{-st}\cdot 0 dt \\
\amp = 7\int_0^3 e^{-st}dt + 0 \\
\amp = 7 \cdot \left[\dfrac{1}{-s}e^{-st}\right]_0^3 \\
\amp = 7 \cdot \left[\left(\dfrac{1}{-s}e^{-s\cdot 3}\right) - \left(\dfrac{1}{-s}e^{-s\cdot 0}\right)\right] \\
\amp = 7 \cdot \left[\left(\dfrac{1}{-s}e^{-s\cdot 3}\right) - \left(\dfrac{1}{-s}\cdot 1\right)\right] \\
\amp = 7 \cdot \left[-\dfrac{1}{s}e^{-3s} +\dfrac{1}{s}\cdot 1\right] \\
\amp = 7 \cdot \left[-\dfrac{1}{s}e^{-3s} +\dfrac{1}{s}\right] \\
\amp = \dfrac{7}{s} - \dfrac{7}{s}e^{-3s}
\end{align*}
14 First-Order Linear Systems
14.8 Exercises
✍🏻 Problems
14.8.1.
Solution.
We can take the LT of each DE. The result is two algebraic equations in two unknowns \(X(s) \) and \(Y(s). \) We will solve the equations simultaneously as shown below.
\begin{align*}
\frac{dx}{dt} \amp = -x+y \amp \frac{dy}{dt} \amp = 2x \\
\lap{\frac{dx}{dt}} \amp = \lap{-x+y} \amp \lap{\frac{dy}{dt}} \amp = \lap{2x} \\
sX(s) - x(0) \amp = -\lap{x} + \lap{y} \amp sY(s) - y(0) \amp = 2 \lap{x} \\
sX(s) - 0 \amp = -X(s) + Y(s) \amp sY(s) - 1 \amp = 2X(s) \\
Y(s) \amp = sX(s) + X(s) \amp \amp \\
\amp \amp s\Big[ sX(s) + X(s) \Big] - 1 \amp = 2X(s) \\
\amp \amp s^2 X(s) + sX(s) - 1 \amp = 2X(s) \\
\amp \amp s^2X(s) + sX(s) - 2X(s) \amp = 1 \\
\amp \amp X(s) [s^2 + s - 2] \amp = 1 \\
\amp \amp X(s) \amp = \frac{1}{s^2 + s - 2} \\
\amp \amp \amp = \frac{1}{(s+2)(s-1)} \\
Y(s) \amp = sX(s) + X(s) \amp \amp \\
\amp = X(s)[s+1] \amp \amp \\
\amp = \frac{1}{(s+2)(s-1)}\cdot (s+1) \amp \amp \\
\amp = \frac{s+1}{(s+2)(s-1)} \amp \amp
\end{align*}
We need only take the inverse LT of each function in order to solve for the desired function \(x(t) \) and \(y(t). \) This means we will need to find a partial fraction decomposition for each.
\begin{align*}
\frac{1}{(s+2)(s-1)} \amp = \frac{A}{s+2} + \frac{B}{s-1} \\
1 \amp = A(s-1) + B(s+2) \\
0s + 1 \amp = As - A + Bs + 2B \\
\amp = (A+B)s + (-A + 2B)
\end{align*}
\begin{align*}
A+B \amp = 0 \amp -A + 2B \amp = 1 \\
B \amp = -A \amp \amp \\
\amp \amp -A + 2(-A) \amp = 1 \\
\amp \amp -3A \amp = 1 \\
\amp \amp A \amp = -\frac{1}{3} \\
B \amp = -\left( -\frac{1}{3} \right)\amp \amp \\
\amp = \frac{1}{3}\amp \amp
\end{align*}
Hence,
\begin{align*}
X(s) \amp = \frac{-\frac{1}{3}}{s+2} + \frac{\frac{1}{3}}{s-1}\\
\amp = \frac{1}{3}\left[ \frac{1}{s-1} - \frac{1}{s+2} \right]
\end{align*}
Similarly, we will find a partial fraction decomposition for\(Y(s). \)
\begin{align*}
\frac{s+1}{(s+2)(s-1)} \amp = \frac{A}{s+2} + \frac{B}{s-1} \\
s + 1 \amp = A(s-1) + B(s+2) \\
\amp = As - A + Bs + 2B \\
\amp = (A+B)s + (-A + 2B)
\end{align*}
\begin{align*}
A+B \amp = 1 \amp -A + 2B \amp = 1 \\
B \amp = 1-A \amp \amp \\
\amp \amp -A + 2(1-A) \amp = 1 \\
\amp \amp -3A + 2 \amp = 1 \\
\amp \amp -3A \amp = -1 \\
\amp \amp A \amp = \frac{1}{3} \\
B \amp = 1-\left( \frac{1}{3} \right)\amp \amp \\
\amp = \frac{2}{3}\amp \amp
\end{align*}
Hence,
\begin{align*}
Y(s) \amp = \frac{\frac{1}{3}}{s+2} + \frac{\frac{2}{3}}{s-1}\\
\amp = \frac{1}{3}\left[ \frac{1}{s+2} + 2\cdot \frac{1}{s-1} \right]
\end{align*}
Now we need only find the inverse LT of equations.
\begin{align*}
x(t) \amp = \lap^{-1}\left\{ X(s) \right\} \\
\amp = \lap^{-1}\left\{ \frac{1}{3}\left[ \frac{1}{s-1} - \frac{1}{s+2} \right] \right\} \\
\amp = \frac{1}{3}\left[ \lap^{-1}\left\{\frac{1}{s-1}\right\} - \lap^{-1}\left\{\frac{1}{s+2}\right\} \right] \\
\amp = \frac{1}{3}\left[ e^t - e^{-2t} \right] \\
y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\
\amp = \lap^{-1}\left\{ \frac{1}{3}\left[ \frac{1}{s+2} +2\cdot\frac{1}{s-1} \right] \right\} \\
\amp = \frac{1}{3}\left[ \lap^{-1}\left\{\frac{1}{s+2}\right\} +2 \lap^{-1}\left\{\frac{1}{s-1}\right\} \right] \\
\amp = \frac{1}{3}\left[ e^{-2t} + 2e^{t} \right]
\end{align*}
This, the solution to this system is
\begin{align*}
x(t) \amp = \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \\
y(t) \amp = \frac{1}{3}e^{-2t} + \frac{2}{3}e^t
\end{align*}
We can verify our solution.
\begin{align*}
\mbox{LHS of first DE} \amp = \frac{dx}{dt} \\
\amp = \frac{d}{dt}\left( \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \right) \\
\amp = \frac{1}{3}e^{t} + \frac{2}{3}e^{-2t} \\
\mbox{RHS of first DE} \amp = -x+y \\
\amp = -\left( \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \right) + \left( \frac{1}{3}e^{-2t} + \frac{2}{3}e^t \right) \\
\amp = \frac{1}{3}e^t + \frac{2}{3}e^{-2t} \\
\amp = \mbox{LHS of first DE} \\
\mbox{LHS of second DE} \amp = \frac{dy}{dt} \\
\amp = \frac{d}{dt}\left( \frac{1}{3}e^{-2t} + \frac{2}{3}e^t \right) \\
\amp = -\frac{2}{3}e^{-2t} + \frac{2}{3}e^t \\
\mbox{RHS of second DE} \amp = 2x \\
\amp = 2\left( \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \right) \\
\amp = \frac{2}{3}e^t - \frac{2}{3}e^{-2t} \\
\amp = \mbox{LHS of second DE} \\
x(0) \amp = \frac{1}{3}e^0 - \frac{1}{3}e^{-2\cdot 0} \\
\amp = \frac{1}{3}\cdot 1 - \frac{1}{3}\cdot 1 \\
\amp = 0 \\
y(0) \amp = \frac{1}{3}e^{-2\cdot 0} + \frac{2}{3}e^0 \\
\amp = \frac{1}{3}\cdot 1 + \frac{2}{3}\cdot 1 \\
\amp = 1
\end{align*}
Hence, the solution satisfies both DEs and both initial conditions.
14.8.2.
Solution.
We can take the LT of each DE, starting with the first DE. We will then solve for\(X(s). \)
\begin{align*}
\frac{dx}{dt} \amp = x-2y\\
\lap{\frac{dx}{dt}} \amp = \lap{x-2y}\\
sX(s) - x(0) \amp = \lap{x} - 2 \lap{y}\\
sX(s) + 1 \amp = X(s) - 2Y(s)\\
sX(s) - X(s) \amp = -2Y(s) - 1\\
X(s)[s-1] \amp = -2Y(s) - 1\\
X(s) \amp = \frac{-2}{s-1}Y(s) - \frac{1}{s-1}
\end{align*}
Now we take the LT of the second DE. We will substitute in for\(X(s) \) using the result we found above.
\begin{align*}
\frac{dy}{dt} \amp = 5x - y \\
\lap{\frac{dy}{dt}} \amp = \lap{5x - y} \\
sY(s) - y(0) \amp = 5\lap{x} - \lap{y} \\
sY(s) - 2 \amp = 5X(s) - Y(s) \\
sY(s) - 2 \amp = 5\left[ \frac{-2}{s-1}Y(s) - \frac{1}{s-1} \right] - Y(s) \\
sY(s) - 2 \amp = \frac{-10}{s-1}Y(s) - \frac{5}{s-1}- Y(s) \\
(s-1) \cdot \left( sY(s) - 2 \right) \amp = \left( \frac{-10}{s-1}Y(s) - \frac{5}{s-1}- Y(s) \right) \cdot (s-1) \\
(s^2 - s)Y(s) - 2s + 2 \amp = -10Y(s) - 5 - (s-1)Y(s) \\
(s^2 - s)Y(s) + 10Y(s) + (s-1)Y(s) \amp = 2s - 2 - 5 \\
Y(s)[s^2 - s + 10 + s - 1] \amp = 2s - 7 \\
Y(s)[s^2 + 9] \amp = 2s - 7 \\
Y(s) \amp = \frac{2s - 7}{s^2 + 9}
\end{align*}
We now substitute this back into equation.
\begin{align*}
X(s) \amp = \frac{-2}{s-1}Y(s) - \frac{1}{s-1} \\
\amp = \frac{-2}{s-1}\cdot\frac{2s - 7}{s^2 + 9} - \frac{1}{s-1} \\
\amp = \frac{-4s+14}{(s-1)(s^2 + 9)} - \frac{1}{s-1}\cdot \frac{s^2 + 9}{s^2 + 9} \\
\amp = \frac{(-4s + 14) - (s^2 + 9) }{(s-1)(s^2 + 9)} \\
\amp = \frac{-4s + 14 - s^2 - 9) }{(s-1)(s^2 + 9)} \\
\amp = \frac{-s^2 - 4s + 5}{(s-1)(s^2 + 9)} \\
\amp = \frac{-(s^2 + 4s - 5)}{(s-1)(s^2 + 9)} \\
\amp = \frac{-(s+5)(s-1)}{(s-1)(s^2 + 9)} \\
\amp = \frac{-(s+5)}{s^2 + 9} \\
\amp = \frac{-s-5}{s^2 + 9}
\end{align*}
We need only take the inverse LT of each function in order to solve for the desired functions \(x(t) \) and \(y(t). \)
\begin{align*}
x(t) \amp = \lap^{-1}\left\{ X(s) \right\} \\
\amp = \lap^{-1}\left\{ \frac{-s-5}{s^2 + 9} \right\} \\
\amp = - \lap^{-1}\left\{ \frac{s}{s^2 + 9} \right\} - 5 \lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\} \\
\amp = - \cos(3t) - 5 \lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\}\cdot \frac{3}{3} \\
\amp = - \cos(3t) - 5 \lap^{-1}\left\{ \frac{3}{s^2 + 3^2} \right\}\cdot \frac{1}{3} \\
\amp = - \cos(3t) - \frac{5}{3} \sin(3t)
\end{align*}
\begin{align*}
y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\
\amp = \lap^{-1}\left\{ \frac{2s - 7}{s^2 + 9} \right\} \\
\amp = 2\lap^{-1}\left\{ \frac{s}{s^2 + 9} \right\} - 7\lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\} \\
\amp = 2\cos(3t) - 7\lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\}\cdot \frac{3}{3} \\
\amp = 2\cos(3t) - 7\lap^{-1}\left\{ \frac{3}{s^2 + 3^2} \right\}\cdot \frac{1}{3} \\
\amp = 2\cos(3t) - \frac{7}{3}\sin(3t)
\end{align*}
Thus, the solution to this system is
\begin{align*}
x(t) \amp = -\cos(3t) - \frac{5}{3} \sin(3t) \\
y(t) \amp = 2\cos(3t) - \frac{7}{3}\sin(3t)
\end{align*}
We can verify our solution.
\begin{align*}
\mbox{LHS of first DE} \amp = \frac{dx}{dt} \\
\amp = \frac{d}{dt}\left( -\cos(3t) - \frac{5}{3} \sin(3t) \right) \\
\amp = 3\sin(3t) - 5\cos(3t) \\
\mbox{RHS of first DE} \amp = x-2y \\
\amp = \left( -\cos(3t) - \frac{5}{3} \sin(3t) \right) -2 \left( 2\cos(3t) - \frac{7}{3}\sin(3t) \right) \\
\amp = -\cos(3t) - \frac{5}{3} \sin(3t) - 4\cos(3t) + \frac{14}{3}\sin(3t) \\
\amp = -5\cos(3t) + \frac{9}{3}\sin(3t) \\
\amp = -5\cos(3t) + 3\sin(3t) \\
\amp = \mbox{LHS of first DE} \\
\mbox{LHS of second DE} \amp = \frac{dy}{dt} \\
\amp = \frac{d}{dt}\left( 2\cos(3t) - \frac{7}{3}\sin(3t) \right) \\
\amp = -6\sin(3t) - 7\cos(3t) \\
\mbox{RHS of second DE} \amp = 5x - y \\
\amp = 5\left( -\cos(3t) - \frac{5}{3} \sin(3t) \right) - \left( 2\cos(3t) - \frac{7}{3}\sin(3t) \right) \\
\amp = -5\cos(3t) - \frac{25}{3} \sin(3t) - 2\cos(3t) + \frac{7}{3}\sin(3t) \\
\amp = -7\cos(3t)- \frac{18}{3} \sin(3t) \\
\amp = -7\cos(3t) - 6\sin(3t) \\
\amp = \mbox{LHS of second DE} \\
x(0) \amp = -\cos(3\cdot 0) - \frac{5}{3} \sin(3\cdot 0) \\
\amp = -1 - 0 \\
\amp = -1 \\
y(0) \amp = 2\cos(3\cdot 0) - \frac{7}{3}\sin(3\cdot 0) \\
\amp = 2\cdot 1 - 0 \\
\amp = 2
\end{align*}
Hence, the solution satisfies both DEs and both initial conditions.
14.8.3.
Solution.
We can take the LT of each DE, starting with the first DE. We will then solve for\(X(s). \)
\begin{align*}
y' - 2x \amp = 1\\
\lap{ y' - 2x } \amp = \lap{ 1 }\\
sY(s) -y(0) - 2X(s) \amp = \frac{1}{s}\\
sY(s) - 0 - 2X(s) \amp = \frac{1}{s}\\
sY(s) \amp = \frac{1}{s}+ 2X(s)\\
Y(s) \amp = \frac{1}{s^2} + \frac{2}{s}X(s)
\end{align*}
Now we take the LT of the second DE. We will substitute in for\(Y(s) \) using the result we found above.
\begin{align*}
x' + y' - 3x - 3y \amp = 2 \\
\lap{ x' + y' - 3x - 3y } \amp = \lap{2} \\
sX(s) - x(0) + sY(s) - y(0) - 3X(s) - 3Y(s) \amp = \frac{2}{s} \\
sX(s) - 0 + sY(s) - 0 - 3X(s) - 3Y(s) \amp = \frac{2}{s} \\
sX(s) - 3X(s) + (s-3) Y(s) \amp = \frac{2}{s} \\
sX(s) - 3X(s) + (s-3)\left[ \frac{1}{s^2} + \frac{2}{s}X(s) \right] \amp = \frac{2}{s} \\
sX(s) - 3X(s) + \frac{s-3}{s^2} + \frac{2(s-3)}{s}X(s) \amp = \frac{2}{s} \\
sX(s) - 3X(s) + \frac{2(s-3)}{s}X(s) \amp = \frac{2}{s} - \frac{s-3}{s^2} \\
s^2 \cdot \left[sX(s) - 3X(s) + \frac{2(s-3)}{s}X(s)\right] \amp = s^2\cdot \left[\frac{2}{s} - \frac{s-3}{s^2}\right] \\
s^3X(s) - 3s^2X(s) + 2s(s-3)X(s) \amp = 2s - (s-3) \\
X(s) [s^3 - 3s^2 + (2s^2 - 6s)] \amp = 2s - s + 3 \\
X(s) [s^3 - s^2 - 6s] \amp = s+3 \\
X(s) \amp = \frac{s+3}{s^3 - s^2 - 6s} \\
\amp = \frac{s+3}{s(s^2 - s - 6)} \\
\amp = \frac{s+3}{s(s-3)(s+2)}
\end{align*}
We now substitute this back into equation.
\begin{align*}
Y(s) \amp = \frac{1}{s^2} + \frac{2}{s}X(s) \\
\amp = \frac{1}{s^2} + \frac{2}{s}\cdot \frac{s+3}{s(s-3)(s+2)} \\
\amp = \frac{1}{s^2} + \frac{2(s+3)}{s^2(s-3)(s+2)} \\
\amp = \frac{1}{s^2}\cdot \frac{(s-3)(s+2)}{(s-3)(s+2)} + \frac{2(s+3)}{s^2(s-3)(s+2)} \\
\amp = \frac{(s-3)(s+2) + 2(s+3)}{s^2(s-3)(s+2)} \\
\amp = \frac{s^2 - s - 6 + 2s + 6}{s^2(s-3)(s+2)} \\
\amp = \frac{s^2 + s}{s^2(s-3)(s+2)} \\
\amp = \frac{s(s+1)}{s^2(s-3)(s+2)} \\
\amp = \frac{s+1}{s(s-3)(s+2)}
\end{align*}
We need only take the inverse LT of each function in order to solve for the desired function \(x(t) \) and \(y(t). \) This means we will need to find a partial fraction decomposition for each.
\begin{align*}
\frac{s+3}{s(s-3)(s+2)} \amp = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+2} \\
s+3 \amp = A(s-3)(s+2) + B(s)(s+2) + C(s)(s-3) \\
0s^2 + s + 3 \amp = A(s^2 - s - 6) + B(s^2 + 2s) + C(s^2 - 3s) \\
\amp = As^2 - As - 6A + Bs^2 + 2Bs + Cs^2 - 3Cs \\
\amp = (A+B+C)s^2 + (-A + 2B- 3C)s + (-6A)
\end{align*}
\begin{align*}
A+B+C \amp = 0 \amp -A + 2B - 3C \amp = 1 \amp -6A \amp = 3 \\
\amp \amp \amp \amp A \amp = -\frac{1}{2} \\
B + C \amp = -A \amp 2B - 3C \amp = 1 + A \amp \amp \\
B + C \amp = -\left( -\frac{1}{2} \right) \amp 2B - 3C \amp = 1 + \left( -\frac{1}{2} \right) \amp \amp \\
B + C \amp = \frac{1}{2} \amp 2B - 3C \amp = \frac{1}{2} \amp \amp \\
C \amp = \frac{1}{2} - B \amp \amp \amp \amp \\
\amp \amp 2B - 3\left( \frac{1}{2} - B \right) \amp = \frac{1}{2} \amp \amp \\
\amp \amp 2B - \frac{3}{2} + 3B \amp = \frac{1}{2} \amp \amp \\
\amp \amp 5B \amp = \frac{1}{2} + \frac{3}{2} \amp \amp \\
\amp \amp 5B \amp = 2 \amp \amp \\
\amp \amp B \amp = \frac{2}{5} \amp \amp \\
C \amp = \frac{1}{2} - B \amp \amp \amp \amp \\
\amp = \frac{1}{2} - \frac{2}{5} \amp \amp \amp \amp \\
\amp = \frac{1}{10} \amp \amp \amp \amp
\end{align*}
Hence,
\begin{align*}
X(s) \amp = \frac{-\frac{1}{2}}{s} + \frac{\frac{2}{5}}{s-3} + \frac{\frac{1}{10}}{s+2}\\
\amp = -\frac{1}{2}\cdot \frac{1}{s} + \frac{2}{5}\cdot \frac{1}{s-3} + \frac{1}{10}\cdot \frac{1}{s+2}
\end{align*}
Similarly, we will find a partial fraction decomposition for\(Y(s). \)
\begin{align*}
\frac{s+1}{s(s-3)(s+2)} \amp = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+2} \\
s+1 \amp = A(s-3)(s+2) + B(s)(s+2) + C(s)(s-3) \\
0s^2 + s + 1 \amp = A(s^2 - s - 6) + B(s^2 + 2s) + C(s^2 - 3s) \\
\amp = As^2 - As - 6A + Bs^2 + 2Bs + Cs^2 - 3Cs \\
\amp = (A+B+C)s^2 + (-A + 2B- 3C)s + (-6A)
\end{align*}
\begin{align*}
A+B+C \amp = 0 \amp -A + 2B - 3C \amp = 1 \amp -6A \amp = 1 \\
\amp \amp \amp \amp A \amp = -\frac{1}{6} \\
B + C \amp = -A \amp 2B - 3C \amp = 1 + A \amp \amp \\
B + C \amp = -\left( -\frac{1}{6} \right) \amp 2B - 3C \amp = 1 + \left( -\frac{1}{6} \right) \amp \amp \\
B + C \amp = \frac{1}{6} \amp 2B - 3C \amp = \frac{5}{6} \amp \amp \\
C \amp = \frac{1}{6} - B \amp \amp \amp \amp \\
\amp \amp 2B - 3\left( \frac{1}{6} - B \right) \amp = \frac{5}{6} \amp \amp \\
\amp \amp 2B - \frac{1}{2} + 3B \amp = \frac{5}{6} \amp \amp \\
\amp \amp 5B \amp = \frac{5}{6} + \frac{1}{2} \amp \amp \\
\amp \amp 5B \amp = \frac{4}{3} \amp \amp \\
\amp \amp B \amp = \frac{4}{15} \amp \amp \\
C \amp = \frac{1}{6} - B \amp \amp \amp \amp \\
\amp = \frac{1}{6} - \frac{4}{15}\amp \amp \amp \amp \\
\amp = \frac{5}{30} - \frac{8}{30}\amp \amp \amp \amp \\
\amp = -\frac{1}{10} \amp \amp \amp \amp
\end{align*}
Hence,
\begin{align*}
Y(s) \amp = \frac{-\frac{1}{6}}{s} + \frac{\frac{4}{15}}{s-3} + \frac{-\frac{1}{10}}{s+2}\\
\amp = -\frac{1}{6}\cdot \frac{1}{s} + \frac{4}{15}\cdot \frac{1}{s-3} - \frac{1}{10}\cdot \frac{1}{s+2}
\end{align*}
Now we need only find the inverse LT of equations.
\begin{align*}
x(t) \amp = \lap^{-1}\left\{ X(s) \right\} \\
\amp = \lap^{-1}\left\{ -\frac{1}{2}\cdot \frac{1}{s} + \frac{2}{5}\cdot \frac{1}{s-3} + \frac{1}{10}\cdot \frac{1}{s+2} \right\} \\
\amp = -\frac{1}{2}\lap^{-1}\left\{ \frac{1}{s} \right\} + \frac{2}{5} \lap^{-1}\left\{ \frac{1}{s-3} \right\} + \frac{1}{10}\lap^{-1}\left\{ \frac{1}{s+2} \right\} \\
\amp = -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \\
y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\
\amp = \lap^{-1}\left\{ -\frac{1}{6}\cdot \frac{1}{s} + \frac{4}{15}\cdot \frac{1}{s-3} - \frac{1}{10}\cdot \frac{1}{s+2} \right\} \\
\amp = -\frac{1}{6}\lap^{-1}\left\{ \frac{1}{s} \right\} + \frac{4}{15}\lap^{-1}\left\{ \frac{1}{s-3} \right\} - \frac{1}{10} \lap^{-1}\left\{ \frac{1}{s+2} \right\} \\
\amp = -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t}
\end{align*}
Thus, the solution to this system is
\begin{align*}
x(t) \amp = -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \\
y(t) \amp = -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t}
\end{align*}
We can verify that this is the solution.
\begin{align*}
\mbox{LHS of first DE} \amp = y' - 2x \\
\amp = \frac{d}{dt}\left( -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \right) - 2\Big( -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \Big) \\
\amp = \left( 0 + \frac{4}{5}e^{3t} + \frac{1}{5} e^{-2t} \right) + 1 - \frac{4}{5}e^{3t} - \frac{1}{5}e^{-2t} \\
\amp = 1 \\
\amp = \mbox{RHS of first DE} \\
\mbox{RHS of second DE} \amp = x' + y' - 3x - 3y \\
\amp = \frac{d}{dt}\left( -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \right) + \frac{d}{dt}\left( -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \right) \\
\amp \mbox{}\hspace{1cm} - 3 \left( -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \right) - 3 \left( -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \right) \\
\amp = \left(0 + \frac{6}{5}e^{3t} - \frac{1}{5}e^{-2t} \right) + \left( 0 + \frac{4}{5}e^{3t} + \frac{1}{5} e^{-2t} \right) \\
\amp \mbox{}\hspace{1cm} + \frac{3}{2} - \frac{6}{5}e^{3t} - \frac{3}{10}e^{-2t} + \frac{1}{2} - \frac{4}{5}e^{3t} + \frac{3}{10} e^{-2t} \\
\amp = \left( \frac{6}{5} + \frac{4}{5} - \frac{6}{5} - \frac{4}{5} \right)e^{3t} + \left( - \frac{1}{5} + \frac{1}{5} - \frac{3}{10} + \frac{3}{10} \right)e^{2t} + \left( \frac{3}{2} + \frac{1}{2} \right) \\
\amp = 0 + 0 + 2 \\
\amp = 2 \\
\amp = \mbox{RHS of second DE} \\
x(0) \amp = -\frac{1}{2} + \frac{2}{5}e^{3\cdot 0} + \frac{1}{10}e^{-2\cdot 0} \\
\amp = -\frac{1}{2} + \frac{2}{5} + \frac{1}{10} \\
\amp = -\frac{5}{10} + \frac{4}{10} + \frac{1}{10} \\
\amp = 0 \\
y(0) \amp = -\frac{1}{6} + \frac{4}{15}e^{3\cdot 0} - \frac{1}{10} e^{-2\cdot 0} \\
\amp = -\frac{1}{6} + \frac{4}{15} - \frac{1}{10} \\
\amp = -\frac{5}{30} + \frac{8}{30} - \frac{3}{30} \\
\amp = 0
\end{align*}
Hence, the solution satisfies both DEs and both initial conditions.
14.8.5.
14.8.6. From a Second-Order Equation to a System.
14.8.7.
Solution.
We consider the rate of change in the amount of salt in tank 1.
\begin{align*}
\frac{dx_1}{dt} \amp = \mbox{(rate of salt in to the tank)} - \mbox{(rate of salt in to the tank)} \\
\amp = \Bigg[\underbrace{0.1 \cdot 4}_{ \frac{\mbox{kg}}{\mbox{L}}\cdot\frac{\mbox{L}}{\mbox{min}}} + \underbrace{\frac{x_2}{100}\cdot 1}_{ \frac{\mbox{kg}}{\mbox{L}}\cdot\frac{\mbox{L}}{\mbox{min}}}\Bigg] - \Bigg[\underbrace{\frac{x_1}{100}\cdot 3}_{ \frac{\mbox{kg}}{\mbox{L}}\cdot\frac{\mbox{L}}{\mbox{min}}}\Bigg] \\
\amp = 0.4 + \frac{1}{100}x_2 - \frac{1}{20}x_1
\end{align*}
Similarly, we consider the rate of change in the amount of salt in tank 2.
\begin{align*}
\frac{dx_2}{dt} \amp = \mbox{(rate of salt in to the tank)} - \mbox{(rate of salt in to the tank)} \\
\amp = \Bigg[\underbrace{\frac{x_1}{100}\cdot 3}_{ \frac{\mbox{kg}}{\mbox{L}}\cdot\frac{\mbox{L}}{\mbox{min}}}\Bigg] - \Bigg[\underbrace{\frac{x_2}{100}\cdot 3}_{ \frac{\mbox{kg}}{\mbox{L}}\cdot\frac{\mbox{L}}{\mbox{min}}}\Bigg] \\
\amp = \frac{3}{100}x_1 - \frac{3}{100}x_2
\end{align*}
Hence, our system is:
\begin{align*}
x'_1 \amp = - \frac{1}{20}x_1+ \frac{1}{100}x_2 + 0.4, \\
x'_2 \amp = \frac{3}{100}x_1 - \frac{3}{100}x_2.
\end{align*}
14.8.8.
14.8.9.
14.8.12.
Solution.
Let \(y(t)\) be the amount of salt in the tank at time \(t\). The rate of change of salt with respect to time, \(\frac{dy}{dt}\), is given by the rate of salt coming in minus the rate of salt going out. The rate of salt coming in is 5 grams/minute. The concentration of salt in the tank is \(\frac{y(t)}{100 - 2t}\) and thus the rate at which salt is removed from the tank is \(2 \times \frac{y(t)}{100 - 2t}\). Therefore, the differential equation is:
\begin{equation*}
\frac{dy}{dt} \amp = 5 - 2 \frac{y(t)}{100 - 2t}
\end{equation*}
14.8.14. Euler’s Method for Systems.
Solution.
\begin{align*}
\mbox{Preliminaries: } \amp \amp f_1(t,x,y) \amp = 2x - y + t \\
\amp \amp f_2(t,x,y) \amp = x \\
\amp \amp h \amp = 0.1 \\
\amp \amp t_0 \amp = 0 \\
\amp \amp x_0 \amp = 6 \\
\amp \amp y_0 \amp = 2 \\
\amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + h \\
\amp \amp \amp = 0 + 0.1 \\
\amp \amp \amp = 0.1 \\
\amp \amp x_1 \amp = x_0 + h \cdot f_1(t_0,x_0,y_0) \\
\amp \amp \amp = 6 + 0.1 \cdot f_1(0,6,2) \\
\amp \amp \amp = 6 + 0.1 \cdot [2(6) - 2 + 0] \\
\amp \amp \amp = 7 \\
\amp \amp y_1 \amp = y_0 + h\cdot f_2(t_0,x_0,y_0) \\
\amp \amp \amp = 2 + 0.1 \cdot f_2(0,6,2) \\
\amp \amp \amp = 2 + 0.1 \cdot [6] \\
\amp \amp \amp = 2.6 \\
\amp \amp \amp \mbox{Iteration 2:} \amp \amp t_2 \amp = t_1 + h \\
\amp \amp \amp = 0.1 + 0.1 \\
\amp \amp \amp = 0.2 \\
\amp \amp x_2 \amp = x_1 + h \cdot f_1(t_1,x_1,y_1) \\
\amp \amp \amp = 7 + 0.1 \cdot f_1(0.1,7,2.6) \\
\amp \amp \amp = 7 + 0.1 \cdot [2(7) - 2.6 + 0.1] \\
\amp \amp \amp = 8.15 \\
\amp \amp y_2 \amp = y_1 + h\cdot f_2(t_1,x_1,y_1) \\
\amp \amp \amp = 2.6 + 0.1 \cdot f_2(0.1,7,2.6) \\
\amp \amp \amp = 2.6 + 0.1 \cdot [7] \\
\amp \amp \amp = 3.3
\end{align*}
Therefore \(x(0.2) \approx 8.15 \) and \(y(0.2) \approx 3.3. \)
14.8.17.
Solution.
We have one second-order and one first-order DE, which means we will need three variables to generate a systemof three first-order DEs.\\ Let \(u = y \) \(v = y', \) and \(w = x \) . Then we have the following.
\begin{align*}
u' \amp = y' = v \label{eq14-9}
\end{align*}
Substituting into the first DE yields the following.
\begin{align*}
y'' + x \amp = 12 \nonumber \\
v' + w \amp = 12 \nonumber \\
v' \amp = 12 - w \label{eq14-10}
\end{align*}
Substituting into the second DE yields the following.
\begin{align*}
w' + 2ut \amp = \cos t \nonumber \\
w' \amp = \cos t - 2ut \label{eq14-11}
\end{align*}
Then equations (\ref{eq14-9}), (\ref{eq14-10}) and (\ref{eq14-11}) constitute a system of first-order DEs in the variable \(u \) \(v \) , and \(w \) , as below:
\begin{align*}
u' \amp = v \\
v' \amp = 12 - w \\
w' \amp = \cos t - 2ut
\end{align*}
14.8.18. Euler’s Method for Higher Order Systems.
Solution.
First, we convert the second-order DE to a system of two first-order DEs. Let \(u = y \) and \(v = y' \text{.}\) Then
\begin{equation}
u' = y' = v\tag{14.2}
\end{equation}
and
\begin{equation}
v' = (y')' = y''.\tag{14.3}
\end{equation}
\begin{align}
y'' + 2y' + y^2 \amp = 0 \tag{14.4}\\
v' + 2v + u^2 \amp = 0 \tag{14.5}\\
v' \amp = -2v - u^2\tag{14.6}
\end{align}
Then equations (14.2) and (14.6) constitute a system of first-order DEs in the variable \(u\) and \(v\text{.}\)
\begin{align*}
u' \amp = v \\
v' \amp = -2v - u^2
\end{align*}
Note that we can also convert the initial conditions to the new variables:
\begin{align*}
u(0) \amp = 1 \\
v(0) \amp = 2
\end{align*}
Now we can use Euler’s method as follows.
\begin{align*}
\mbox{Preliminaries: } \amp \amp f_1(t,u,v) \amp = v \\
\amp \amp f_2(t,u,v) \amp = -2v-u^2 \\
\amp \amp h \amp = 0.1 \\
\amp \amp t_0 \amp = 0 \\
\amp \amp u_0 \amp = 1 \\
\amp \amp v_0 \amp = 2 \\
\amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + h \\
\amp \amp \amp = 0 + 0.1 \\
\amp \amp \amp = 0.1 \\
\amp \amp u_1 \amp = u_0 + h \cdot f_1(t_0,u_0,v_0) \\
\amp \amp \amp = 1 + 0.1 \cdot f_1(0,1,2) \\
\amp \amp \amp = 1 + 0.1 \cdot [2] \\
\amp \amp \amp = 1.2 \\
\amp \amp v_1 \amp = v_0 + h\cdot f_2(t_0,u_0,v_0) \\
\amp \amp \amp = 2 + 0.1 \cdot f_2(0,1,2) \\
\amp \amp \amp = 2 + 0.1 \cdot [-2(2) - (1)^2] \\
\amp \amp \amp = 1.5 \\
\amp \amp \amp \mbox{Iteration 2:} \amp \amp t_2 \amp = t_1 + h \\
\amp \amp \amp = 0.1 + 0.1 \\
\amp \amp \amp = 0.2 \\
\amp \amp u_2 \amp = u_1 + h \cdot f_1(t_1,u_1,v_1) \\
\amp \amp \amp = 1.2 + 0.1 \cdot f_1(0.1, 1.2, 1.5) \\
\amp \amp \amp = 1.2 + 0.1 \cdot [1.5] \\
\amp \amp \amp = 1.35 \\
\amp \amp v_2 \amp = v_1 + h\cdot f_2(t_1,u_1,v_1) \\
\amp \amp \amp = 1.5 + 0.1 \cdot f_2(0.1, 1.2, 1.5) \\
\amp \amp \amp = 1.5 + 0.1 \cdot [-2(1.5) - (1.2)^2] \\
\amp \amp \amp = 1.0560
\end{align*}
Therefore \(u(0.2) \approx 1.35 \) and \(v(0.2) \approx 1.0560. \) This means \(y(0.2) \approx 1.35 \) (and, even though we were not asked for it specifically, \(y'(0.2) \approx 1.0560 \)).
14.8.19. Euler’s Method.
Solution.
First, we convert the second-order DE to a system of two first-order DEs. Let \(u = y \) and \(v = y' \text{.}\) Then
\begin{equation}
u' = y' = v\tag{14.7}
\end{equation}
and
\begin{equation}
v' = (y')' = y''.\tag{14.8}
\end{equation}
\begin{align}
y'' - 2y \amp = e^{t-3} \cos t \tag{14.9}\\
v' - 2u \amp = e^{t-3}\cos t \tag{14.10}\\
v' \amp = e^{t-3}\cos t + 2u\tag{14.11}
\end{align}
Then equations (14.7) and (14.11) constitute a system of first-order DEs in the variable \(u\) and \(v\text{:}\)
\begin{align*}
u' \amp = v \\
v' \amp = e^{t-3}\cos t + 2u
\end{align*}
Note that we can also convert the initial conditions to the new variables:
\begin{align*}
u(3) \amp = -1 \\
v(3) \amp = 0
\end{align*}
Now we can use Euler’s method as follows.
\begin{align*}
\mbox{Preliminaries: } \amp \amp f_1(t,u,v) \amp = v \\
\amp \amp f_2(t,u,v) \amp = e^{t-3}\cos t +2u \\
\amp \amp h \amp = 0.2 \\
\amp \amp t_0 \amp = 3 \\
\amp \amp u_0 \amp = -1 \\
\amp \amp v_0 \amp = 0 \\
\amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + h \\
\amp \amp \amp = 3 + 0.2 \\
\amp \amp \amp = 3.2 \\
\amp \amp u_1 \amp = u_0 + h \cdot f_1(t_0,u_0,v_0) \\
\amp \amp \amp = -1 + 0.2 \cdot f_1(3, -1, 0) \\
\amp \amp \amp = -1 + 0.2 \cdot [0] \\
\amp \amp \amp = -1 \\
\amp \amp v_1 \amp = v_0 + h \cdot f_2(t_0,u_0,v_0) \\
\amp \amp \amp = 0 + 0.2 \cdot f_2(3, -1, 0) \\
\amp \amp \amp = 0 + 0.2 \cdot [e^{3-3}\cos(3) + 2(-1)] \\
\amp \amp \amp = -0.5980
\end{align*}
Therefore \(u(3.2) \approx -1 \text{,}\) which means \(y(3.2) \approx -1 \) (and, even though we were not asked for it specifically, we also know \(y'(3.2) \approx -0.5980 \)).
14.8.20. Euler’s Method.
Solution.
We will use the system of DEs developed earlier and use Euler’s method as follows:
\begin{align*}
\mbox{Preliminaries: } \amp \amp f_1(t,u,v) \amp = v \\
\amp \amp f_2(t,u,v) \amp = e^{t-3}\cos t +2u \\
\amp \amp h \amp = 0.1 \\
\amp \amp t_0 \amp = 3 \\
\amp \amp u_0 \amp = -1 \\
\amp \amp v_0 \amp = 0 \\
\amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + h \\
\amp \amp \amp = 3 + 0.1 \\
\amp \amp \amp = 3.1 \\
\amp \amp u_1 \amp = u_0 + h \cdot f_1(t_0,u_0,v_0) \\
\amp \amp \amp = -1 + 0.1 \cdot f_1(3, -1, 0) \\
\amp \amp \amp = -1 + 0.1 \cdot [0] \\
\amp \amp \amp = -1 \\
\amp \amp v_1 \amp = v_0 + h\cdot f_2(t_0,u_0,v_0) \\
\amp \amp \amp = 0 + 0.1 \cdot f_2(3, -1, 0) \\
\amp \amp \amp = 0 + 0.1 \cdot [e^{3-3}\cos(3) + 2(-1)] \\
\amp \amp \amp = -0.2990 \\
\amp \amp \amp \mbox{Iteration 2:} \amp \amp t_2 \amp = t_1 + h \\
\amp \amp \amp = 3.1 + 0.1 \\
\amp \amp \amp = 3.2 \\
\amp \amp u_2 \amp = u_1 + h \cdot f_1(t_1,u_1,v_1) \\
\amp \amp \amp = -1 + 0.1 \cdot f_1(3.1, -1, -0.2990) \\
\amp \amp \amp = -1 + 0.1 \cdot [-0.2990] \\
\amp \amp \amp = -1.0299 \\
\amp \amp v_2 \amp = v_1 + h\cdot f_2(t_1,u_1,v_1) \\
\amp \amp \amp = -0.2990 + 0.1 \cdot f_2(3.1, -1, -0.2990) \\
\amp \amp \amp = -0.2990 + 0.1 \cdot [e^{3.1 - 3}\cos(3.1) + 2(-1)] \\
\amp \amp \amp = -0.6094
\end{align*}
Therefore \(u(3.2) \approx -1.0299 \text{,}\) which means \(y(3.2) \approx -1.0299 \) (and, even though we were not asked for it specifically, we also know \(y'(3.2) \approx -0.6094 \)).
