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Section 3.6 Exercises
Subsection 💡 Conceptual Quiz
Exercises Exercises
📝: Abbreviations.
Differential Equation
Initial Value Problem
1. True or False.
(a)
In general, differential equations have more than one solution.
True.
True. In general, differential equations have infinitely many solutions.
False.
True. In general, differential equations have infinitely many solutions.
(c)
A single initial condition always yields a particular solution?
True.
False. This would be true only if there were a single constant in the general solution. When there are multiple constants, multiple initial conditions are required.
📝 It does, however, limit the possible values of the constants.
False.
False. This would be true only if there were a single constant in the general solution. When there are multiple constants, multiple initial conditions are required.
📝 It does, however, limit the possible values of the constants.
(d)
\(y = x^2 + 3\) is a solution to the differential equation
\begin{equation*}
\frac{dy}{dx} - 3 = 2x\text{.}
\end{equation*}
True
Incorrect. \(y = x^2 + 3\) is not a solution since
\begin{align*}
\frac{dy}{dx} - 3 \amp = 2x \\
\frac{d}{dx}\left[x^2 + 3\right] - 3 \amp = 2x \\
2x - 3 \amp = 2x \quad \leftarrow \text{false}
\end{align*}
False
Correct! \(y = x^2 + 3\) is not a solution since
\begin{align*}
\frac{dy}{dx} - 3 \amp = 2x \\
\frac{d}{dx}\left[x^2 + 3\right] - 3 \amp = 2x \\
2x - 3 \amp = 2x \quad \leftarrow \text{false}
\end{align*}
(e)
The differential equation,
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x\text{,}
\end{equation*}
is an example of an initial-value problem.
True
Incorrect. An initial value problem specifies initial conditions that are not provided in this equation.
False
Correct! The equation is a differential equation without initial conditions; therefore, it is not an initial-value problem.
(f)
Suppose that for some differential equation,
\(\mathcal{F}\) is the family of solutions.
\(y\) is the general solution.
\(y_1\) and \(y_2\) are two different particular solutions.
Select all the true statements.
\(\mathcal{F}\) is the collection of all particular solutions.
Correct! A family of solutions includes all possible particular solutions.
Applying one or more initial conditions to \(y\) gives \(y_2\text{.}\)
Correct! A particular solution is found by applying initial conditions to the general solution.
\(y\) is a member of \(\mathcal{F}\text{.}\)
Incorrect. The general solution is a form that represents the family of solutions, but it is not itself a member of the family.
\(y_1\) & \(y_2\) are particular solutions to the same IVP.
Incorrect. Different particular solutions correspond to different initial conditions, so they cannot both be solutions to the same IVP.
2. Multiple Choice.
(a) Select-the-Best-Answer .
Which statement best describes a
solution to a differential equation?
A function of the dependent variable.
While the solution is a function, it is not a function of the dependent variable.
A function of the independent variable.
Yes, when you solve a differential equation, you are finding a function of the independent variable.
A numerical value.
It is possible for a solution to be a number, but not in general.
A derivative of the dependent variable.
While derivatives of the dependent variable are involved, they do not describe solutions in general.
(b) Select-the-Best-Answer.
Consider the differential equation with a missing right side:
\begin{equation*}
y'' - \frac{4}{x}y' = \fillinmath{XXXXX}\text{.}
\end{equation*}
If \(y = 2x^3\) is a solution to this equation, what must the right side be?
(c) Select-the-Best-Answer.
What is a family of solutions?
A collection of all possible solutions to a differential equation.
Correct! The family of solutions includes every possible particular solution.
The general solution to a differential equation.
Incorrect. The general solution represents a form of the family of solutions, not the entire set.
A single specific solution to a differential equation.
Incorrect. This describes a particular solution, not the family of solutions.
A solution without any constants.
Incorrect. A solution without constants is typically a particular solution, not the entire family.
(d) Select-the-Best-Answer.
What does it mean to “solve” a differential equation?
To find an unknown function that satisfies the equation.
Correct! The goal of solving a differential equation is to find the function that meets the equation’s conditions.
To find the derivative of a function.
Incorrect. While derivatives are involved, the goal is to find the function, not just its derivative.
To identify the constants in an equation.
Incorrect. Identifying constants might be part of the process, but it is not the primary goal.
To determine the independent variable.
Incorrect. The independent variable is usually known; we solve for the dependent variable.
(e) Select-the-Best-Answer.
Which task is fundamentally different from the others?
Solving a differential equation.
Incorrect. Solving a differential equation is closely related to two other tasks on this list.
Finding the general solution to a differential equation.
Incorrect. Finding the general solution is very similar to two other tasks in this list.
Finding a family of solutions to a differential equation.
Incorrect. Finding a family of solutions is very similar to two other tasks in this list.
Verifying a solution to a differential equation.
Correct! Verifying a solution is very different from tasks that seek to find a solution.
3. Matching.
(a) Fill in the Blanks.
The words on the left have been removed from the statements on the right.
Drag each word to the panel that contains the statement it was removed from.
dependent
The solution is represented by the
general
particular
The
independent
The solution depends on the
satisfies
function
For a
4. Short-Answer Questions.
(a) Algebraic vs. Differential Equation Solutions.
Consider the algebraic equation
\begin{equation*}
2x^2 + 3 = 7x\text{.}
\end{equation*}
State what it means for
\(x = 3\) to be a solution to this equation.
Show how you would verify that
\(x = 3\) is a solution.
Do not solve for \(x\text{.}\)
Show how you would verify that
\(x = 4\) is a solution.
Are there any differences in how you verify solutions to differential equations compared to algebraic equations? Explain.
(b) Explain the Difference.
In a few sentences, explain the difference between a general solution, a family of solutions, and a particular solution.
(c) Explain the Significance.
Explain the significance of initial condition(s) as they relate to the particular and general solution of a differential equation.
Solution 1 . (a)
To be a solution to the equation,
\(x = 3\) must satisfy the equation. That is, when we substitute
\(x = 3\) into the equation, the result simplifies to a true statement.
Substituting \(x = 3\) into the equation, we get
\begin{align*}
\os{\text{LHS}}{\overline{2x^2 + 3}} \amp = \os{\text{RHS}}{\overline{7x}} \\
2(3)^2 + 3 \amp = 7(3) \\
21 \amp = 21 \quad
\end{align*}
Since \(21=21\) is an undeniably true statement, we have confirmed that \(x = 3\) is a solution to the equation.
Substituting \(x = 4\) into the equation, we get
\begin{align*}
\os{\text{LHS}}{\overline{2x^2 + 3}} \amp = \os{\text{RHS}}{\overline{7x}} \\
2(4)^2 + 3 \amp = 7(4) \\
35 \amp = 28 \quad
\end{align*}
Since \(35\ne28\) is not true, \(x = 4\) is not a solution to the equation.
The process of verifying solutions to differential equations is exactly the same. However, with differential equations, we have to be more careful about confirming a “true statement”. With numbers, it is easy to see if \(21=21\) is true, but with functions, it’s a bit trickier. We must ensure that the functions are equal for all values of \(x\) (the independent variable). For example, the statement
\begin{equation*}
e^x + \sin^2 x + \cos^2 x = e^x + 1
\end{equation*}
is true since \(\sin^2 x + \cos^2 x = 1\text{.}\) In contrast, the statement
\begin{equation*}
e^x + \sin x + \cos x = e^x + 1
\end{equation*}
is not true since \(\sin x + \cos x \ne 1\) for all values of \(x\text{.}\) It only takes one value of \(x\) to make the statement false. Let’s try a few values of \(x\) to see this.
\begin{equation*}
x = 0:
\end{equation*}
\begin{align*}
e^0 + \sin 0 + \cos 0 \amp = 1 + 1 \\
2 \amp = 2 \quad
\end{align*}
\begin{equation*}
x = \frac{\pi}{2}:
\end{equation*}
\begin{align*}
e^{\pi/2} + \sin \frac{\pi}{2} + \cos \frac{\pi}{2} \amp = e^{\pi/2} + 1 \\
e^{\pi/2} + 1 \amp = e^{\pi/2} + 1 \quad
\end{align*}
\begin{equation*}
x = \pi:
\end{equation*}
\begin{align*}
e^{\pi} + \sin \pi + \cos \pi \amp = e^{\pi} + 1 \\
e^{\pi} - 1 \amp \ne e^{\pi} + 1 \quad
\end{align*}
The fact that the statement is not true for
\(x = \pi\) is enough to show that this is not a true statement and would not correspond to a solution to a differential equation.
Solution 2 . (b)
A general solution represents the form of all possible solutions to a differential equation, typically with one or more arbitrary constants. A family of solutions is an infinite set of solutions, one for each possible combination of constant values in the general solution. A particular solution is a single solution to a differential equation that satisfies the differential equation for specific values of the constants.
Solution 3 . (c)
The initial condition(s) specify one or more points that the graph of the solution must pass through. This often allows one to solve for the constants in the general solution. Therefore, the initial condition(s) act to reduce the family of solutions down to a smaller set of solutions or, ideally, a single particular solution.
Solution 4 . (d)
Integrating both sides gives
\begin{align*}
\int \frac{dy}{dx}\ dx \amp = \int\left(x + y\right)\ dx \\
y + C_1 \amp = \int x\ dx + \int y\ dx \\
y + C_1 \amp = \frac12 x^2 + C_2 + \int y\ dx \\
y - \int y\ dx \amp = \frac12 x^2 + C_2 - C_1
\end{align*}
Without knowing \(y\text{,}\) we cannot simplify \(\int y\ dx\text{.}\) The obstacle is that we cannot combine these \(y\) variables into a single \(y\) on the left-hand side.
Subsection 🏋️♂️ Practice Drills
Exercises Exercises
1. Matching Each Function to the Equation it Satisfies.
Select the Solutions. For each differential equation, select the functions that are solutions to that equation.
2. \(y''-9y = 0\) .
a. \(\ds \,\, y = 3\) b. \(\ds \,\, y = 0\) c. \(\ds \,\, y = e^{3x}\)
d. \(\ds \,\, y = 3x\) e. \(\ds \,\, y = 9e^x\) f. \(\ds \,\, y = x^9\)
g. \(\ds \,\, y = 4e^{3x}\) h. \(\ds \,\, y = e^{-3x}\) i. \(\ds \,\, y = e^{3x} - 2e^{-3x}\)
3. \(y'' - 10y' + 25y = 0\) .
a. \(\ds \,\, y = 0\) b. \(\ds \,\, y = x^2e^{5x}\) c. \(\ds \,\, y = e^{5x}\)
d. \(\ds \,\, y = 5x\) e. \(\ds \,\, y = xe^{5x}\) f. \(\ds \,\, y = 5e^{5x}\)
g. \(\ds \,\, y = 1/25\) h. \(\ds \,\, y = e^{-5x}\) i. \(\ds \,\, y = (1+x)e^{5x}\)
Find the Hidden Right-Hand Side. For each given
\(y(t)\text{,}\) assume it is a solution to the differential equation with a hidden right side. Give the function that must be on the right for
\(y\) to be a solution to the equation.
4. \(\ y(t) = 2e^{-3t}\) .
5. \(\ y(t) = 3\sin(t^2)\) .
Subsection ✍🏻 Problems
Exercises Exercises
Verifying Solutions. For each, verify if the given function,
\(y\text{,}\) is a solution to the differential equation.
1.
\(y = c_1 \sin x + c_2 \cos x\text{,}\) \(\quad \dfrac{d^2y}{dx^2} + y = 0\)
2.
\(y = c_{1}e^{2x} + c_{2}xe^{2x}\text{,}\) \(\quad \dfrac{d^{2}y}{dx^{2}} - 4\dfrac{dy}{dx} + 4y = 0\)
3.
\(y = 3\sin(x^2)\text{,}\) \(\quad y' - xy'' = 12x^3\sin(x^2)\)
4.
\(y = Ce^{-5x^2} + \dfrac{3}{5}\text{,}\) \(\quad y' = \big( 2x \big) \big( 3 - 5y \big)\)
Find the value of \(r\) . Each of the following differential equations has one or more solutions of the form
\(y(t) = e^{rt}\text{,}\) for different values of constant,
\(r\text{.}\) Find all the values of
\(r\) so that
\(y\) is a solution to the equation.
5.
\(\dfrac{d^2 y}{dx^2} + 6 \dfrac{dy}{dx} = -5y\)
6. 7. 8.
Find the value of \(k\) . Each of the following differential equations has one or more solutions of the form
\(y = t^{k}\ (t > 0)\text{,}\) for different values of constant,
\(k\text{.}\) Find all the values of
\(k\) so that
\(y\) is a solution to the equation.
9.
\(3t^2 y^{\prime\prime} = -11t y^{\prime} + 3y\)
10.
\(t^{2}y''- 11ty' + 32y = 0\)
Find the value of \(m\) . Find all the values of
\(m\) that make
\(y\) a solution to the given equation.
11.
\(\left\{
\begin{array}{l}
y=mx^{2};\\
y'=5x
\end{array}
\right.\)
12.
\(\left\{
\begin{array}{l}
y=\sin(mt);\\
y'' + 9y = 0
\end{array}
\right.\)
13.
\(\left\{
\begin{array}{l}
y=e^{3t};\\
\dfrac{d^2y}{dt^2} - 8 \dfrac{dy}{dt} + my = 0
\end{array}
\right.\)
Solutions to Initial-Value Problems.
14. Answer the following.
Show that \(y = \dfrac{\ln x + c}{x}\) is a solution to the differential equation
\begin{equation*}
x^2y' + xy = 1, \qquad y(9)=8\text{.}
\end{equation*}
Find
\(c\) corresponding to the initial condition.
15. Verifying Particular Solutions of Initial-Value Problems Takes Two Steps.
To verify that a function (e.g.,
\(y = -3e^{x^2} + 3\) ) is a particular solution to an initial-value problem, there is an extra step beyond showing that it satisfies the differential equation. What else must this function satisfy?
16.
Show \(y = 3x^2 - 2x + 1\) is a particular solution to
\begin{equation*}
\frac{dy}{dx} = 6x - 2, \qquad y(0) = 1\text{.}
\end{equation*}
17.
Show \(y = -3e^{x^2} + 3\) is a particular solution to
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \qquad y(0) = 0\text{.}
\end{equation*}
18. Find the Correct Solution.
Consider the differential equation
\begin{equation*}
y^\prime = y-y^2, \qquad y(0) = 7
\end{equation*}
and the following three functions
\begin{equation*}
y_1 = c\sin(-x), \quad y_2 = \frac{1}{c+x^2}, \quad y_3 = \frac{1}{1+ce^{-x}}\text{.}
\end{equation*}
Determine which, if any, of
\(y_1, y_2, \) or
\(y_3\) are solutions to this equation.
For any of the solutions found in (a), find the particular solution that corresponds to the initial condition.
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