Section 7.2 General Solutions
In any nonhomogeneous differential equation, the solution should, to some extent, resemble the right-hand side and we can logically guess what function the solution should look like. Yet, hidden within the overall solution could be terms that cancel out each other, affecting only the structure without contributing directly to the right-hand side.
To explore this, consider the following linear nonhomogeneous constant coefficient (LNCC) equation alongside its homogeneous counterpart:
\begin{align}
y'' - 4y' + 3y =\amp\ 9x\tag{32}\\
y'' - 4y' + 3y =\amp\ 0\tag{33}
\end{align}
Let
\(y_h\) be the solution to the homogeneous equation
(33), and
\(y_p\) be the solution to the nonhomogeneous equation
(32), so we have
\begin{align*}
y_p'' - 4y_p' + 3y_p =\amp\ 9x \\
y_h'' - 4y_h' + 3y_h =\amp\ 0
\end{align*}
Now, adding these equations together and rearranging the terms gives us
\begin{align*}
y_h'' - 4y_h' + 3y_h + y_p'' - 4y_p' + 3y_p =\amp\ 9x \\
y_h'' + y_p'' - 4y_h' + 4y_p' + 3y_h + 3y_p =\amp\ 9x \\
(y_h + y_p)'' - 4(y_h + y_p)' + 3(y_h + y_p) =\amp\ 9x
\end{align*}
This shows that not only is
\(y_p\) the solution to the nonhomogeneous equation
(32), but so is
\(y_h+y_p\text{.}\) This happens because the terms in
\(y_h\) simplify to zero and the terms in
\(y_p\) simplify to
\(9x\text{.}\)
From the previous section, we know
\(y_p = 3x + 4\) is a solution to the LNCC equation
(32) and
\(y_h = c_1e^{x} + c_2e^{3x}\) is the solution to the LHCC equation
(33). Therefore, the combined solution
\begin{equation*}
y = y_h + y_p = \ub{c_1e^{x} + c_2e^{3x}}_{\text{simplify to } 0} + \ub{3x + 4}_{\text{simplify to }9x}
\end{equation*}
is a solution to the LNCC equation
(32) where
\(y_h\) contains the terms that simplify to zero, whereas,
\(y_p\) contains the terms that simplify to
\(f(x)\) when
\(y\) is substituted into
(32).
This example leads us to the following concept for the general solution of a LNCC differential equation.
LNCC General Solution Parts.
The linear nonhomogeneous constant coefficient (LNCC) equation
\begin{equation}
a_n y^{(n)} + \cdots + a_1 y' + a_0 y = f(x)\tag{34}
\end{equation}
has a solution with a homogeneous and particular part given by
\begin{equation}
y = y_h + y_p\text{.}\tag{35}
\end{equation}
where \(y_h\text{,}\) found by solving the LHCC equation
\begin{equation*}
a_n y^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = 0\text{.}
\end{equation*}
Here,
\(y_h\) contains the terms of the solution that cancel out, whereas,
\(y_p\) contains the terms that simplify to
\(f(x)\) when
(35) is substituted into
(34).
Example 14.
\(\ \ \)
\begin{equation*}
y'' - 4y' - 12y = 3{e^{5t}}
\end{equation*}
given that the the particular solution is known to be \(y_p(t) = -\frac{3}{7}{{e}^{5t}}\text{.}\)
Solution.
The general solution has the form:
\begin{equation*}
y = y_h + y_p
\end{equation*}
Since \(y_p\) is given, we only need to solve the homogeneous equation:
\begin{equation*}
y'' - 4y' - 12y = 0
\end{equation*}
Using the characteristic equation:
\begin{align*}
r^2 - 4r - 12 =\amp\ 0 \\
(r - 6)(r + 2) =\amp\ 0 \quad \Rightarrow \quad r_1 = 6, r_2 = -2
\end{align*}
The homogeneous solution is:
\begin{equation*}
y_h(t) = c_1e^{-2t} + c_2e^{6t}
\end{equation*}
Therefore, the general solution is:
\begin{equation*}
y = c_1e^{-2t} + c_2e^{6t} -\frac{3}{7}{e^{5t}}
\end{equation*}
Before we can tackle solving these equations from scratch, we need a strategy for finding the particular solution \(y_p\text{,}\) which will be covered in the upcoming sections.
Reading Questions Check-Point Questions
1. If \(y_h = c_1e^{-x} + c_2e^{2x}\) and \(y_p = 5x - 3\text{,}\) what is the general solution to the LNCC equation?
- \(y = c_1e^{-x} + c_2e^{2x} + 5x - 3\text{.}\)
- Correct! The general solution combines the homogeneous and particular parts.
- \(y = c_1e^{-x} + c_2e^{2x} - 5x + 3\text{.}\)
- Incorrect. The signs in the particular solution are wrong.
- \(y = c_1e^{-x} + c_2e^{2x}\text{.}\)
- Incorrect. This is only the homogeneous solution, not the complete general solution.
- \(y = 5x - 3\text{.}\)
- Incorrect. This is only the particular solution, not the full general solution.
2. What is the purpose of the particular solution \(y_p\) in solving a non-homogeneous linear differential equation?
To represent the general solution of the homogeneous equation.
- Incorrect. The general solution of the homogeneous equation is called the complementary solution, not the particular solution.
To determine the coefficients of the characteristic equation.
- Incorrect. The characteristic equation is related to the complementary solution and does not involve the particular solution.
To account for the non-homogeneous term \(f(x)\) on the right-hand side of the equation.
- Correct! The particular solution is chosen to match the form of \(f(x)\) and account for its influence in the equation.
To simplify the process of solving the differential equation.
- Incorrect. The particular solution addresses the specific form of \(f(x)\) and is part of solving the non-homogeneous equation, but its purpose isn’t simplification.
3. How is the homogeneous solution \(y_h\) of an LNCC equation typically found?
- By solving the characteristic equation associated with the homogeneous equation.
- Correct! The characteristic equation provides the exponents for the homogeneous solution.
- By integrating the equation twice.
- Incorrect. Solving the characteristic equation is the standard method for homogeneous solutions.
- By guessing the solution and checking.
- Incorrect. The characteristic equation is the systematic way to find the homogeneous solution.
- By using boundary conditions.
- Incorrect. Boundary conditions are used to find specific constants, not to find \(y_h\text{.}\)
4. Which of the following statements are true about the particular part of the solution, \(y_p\text{,}\) of an LNCC equation?
Which of the following statements are true about the particular part of the solution, \(y_p\text{,}\) of the LNCC equation
\begin{equation}
a_n y^{(n)} + \cdots + a_1 y' + a_0 y = f(x)\tag{36}
\end{equation}
It is part of the general solution of
(36).
- Correct! The general solution is the sum of the homogeneous solution \(y_h\) and the particular solution \(y_p\text{.}\)
It contains the terms in the general solution that simplify to zero when you plug it into the left-side of
(36).
- Incorrect. \(y_h\) is the solution to the homogeneous part, while \(y_p\) is the particular solution.
It contains the terms in the general solution that account for \(f\text{.}\)
- Correct. The terms of \(y_p\) simplify to \(f\) when plugged into (36).
It is found by solving an LHCC equation.
- Incorrect. \(y_h\) is the solution to the homogeneous equation, and \(y_p\) solves the nonhomogeneous equation.
It contains constants of integration.
- Incorrect The homogeneous solution includes constants of integration that are determined by the initial conditions.
It resembles \(f\text{.}\)
- Correct! The particular solution looks like the non-zero right-hand side of the equation.
5. Which of the following statements are true about the homogeneous part of the solution, \(y_h\text{,}\) of an LNCC equation?
Which of the following statements are true about the homogeneous part of the solution,, \(y_p\text{,}\) of the LNCC equation
\begin{equation}
a_n y^{(n)} + \cdots + a_1 y' + a_0 y = f(x)\tag{37}
\end{equation}
It is part of the general solution of
(37)
- Correct! The general solution is the sum of the homogeneous solution \(y_h\) and the particular solution \(y_p\text{.}\)
It contains the terms in the general solution that simplify to zero when you plug it into the left-side of
(37).
- Correct! \(y_h\) is the solution to the homogeneous part, while \(y_p\) is the particular solution.
It contains the terms in the general solution that account for \(f\text{.}\)
- Incorrect. \(y_h\) is related to the homogeneous equation, while the terms of \(y_p\) simplify to \(f\) when plugged into (37).
It is found by solving an LHCC equation.
- Correct! \(y_h\) is the solution to the homogeneous equation, and \(y_p\) solves the nonhomogeneous equation.
It contains constants of integration.
- Correct! The homogeneous solution includes constants of integration that are determined by the initial conditions.
It resembles \(f\text{.}\)
- Incorrect The particular solution accounts for the non-zero right-hand side of the equation.
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