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Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Section 4.3 Exercises

Subsection 💡 Conceptual Quiz

Exercises Exercises

1. True or False.
(a) True-or-False.
We can solve
\begin{equation*} \dfrac{dy}{dx} = x^3 - 7 \end{equation*}
for \(y\) by differentiating both sides with respect to \(x\text{.}\)
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  • True
  • Incorrect, taking a derivative of both sides will result in a second derivative on the left side of the equation.
  • False
  • Correct! We should integrate both sides to solve for \(y\text{,}\) not differentiate.
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(b) True-or-False.
Solving for \(y\) in the equation
\begin{equation*} \dfrac{dy}{dx} = \ln(3x+1) \end{equation*}
amounts to finding the antiderivative of \(\ln(3x+1)\text{.}\)
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  • True
  • Correct, integrating both sides gives
    \begin{equation*} y = \int \ln(3x+1)\ dx \quad \leftarrow \text{antiderivative of } \ln(3x+1)\text{.} \end{equation*}
  • False
  • Incorrect.
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(c) True-or-False.
Combining constants is a common practice in differential equations.
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  • True
  • Correct!
  • False
  • Incorrect, revisit the examples above.
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(d) True-or-False.
    Solving a differential equation by direct integration involves computing a derivative.
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  • True.

  • Direct integration involves integrating both sides of the equation, not computing a derivative.

  • False.

  • Direct integration involves integrating both sides of the equation, not computing a derivative.
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(e) True-or-False.
Direct integration could be used to solve the equation
\begin{equation*} \dfrac{d}{dx}\left[y^2 + x^3\right] = \sqrt{x}\text{.} \end{equation*}
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  • True
  • Correct!
  • False
  • Incorrect. This equation is in the form (0).
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2. Multiple Choice.
(a) Select-the-Best-Answer.
How could you solve for \(y\) in the equation
\begin{equation*} \frac12\dfrac{dy}{dx} - \tan(2x) = x\text{?} \end{equation*}
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  • Differentiating both sides with respect to \(x\text{.}\)
  • Incorrect, differentiating both sides only puts another derivative on \(\dfrac{dy}{dx}\text{.}\)
  • Isolate \(\dfrac{dy}{dx}\) and integrate both sides with respect to \(x\text{.}\)
  • Correct!
  • Isolate \(\dfrac{dy}{dx}\) and integrate both sides with respect to \(y\text{.}\)
  • Incorrect, the integration is not with respect to \(y\text{.}\)
  • Find the antiderivative of \(\tan(2x)\text{.}\)
  • Incorrect, the solution is the antiderivative of \(2\tan(2x) + 2x\text{,}\) not just \(\tan(2x)\text{.}\)
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(b) Select-the-Best-Answer.
The solution to the differential equation
\begin{equation*} \frac13 y' - 7x + x^2 = 1 \end{equation*}
is the antiderivative of which function?
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  • \(\quad y\)
  • Incorrect. \(y\) is the solution to the differential equation.
  • \(\quad 21x - 3x^2 + 1\)
  • Incorrect, perhaps check your algebra.
  • \(\quad 7x - x^2 - 1\)
  • Incorrect, perhaps check your algebra.
  • \(\quad 21x - 3x^2 + 3\)
  • Correct! Isolating \(y'\) gives
    \begin{equation*} y' = 21x - 3x^2 + 3\text{,} \end{equation*}
    so the solution is the antiderivative of \(21x - 3x^2 + 3\text{.}\)
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(c) Select-the-Best-Answer.
Give the reason direct integration cannot be applied to the equation
\begin{equation*} \dfrac{d}{dx}\left[\dfrac{x}{y^2}\right] = \sin(x+y)\text{.} \end{equation*}
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  • There is a fraction in the derivative.
  • The expression in the derivative can be any function of \(x\) and \(y\text{.}\)
  • The \(y\) term is squared.
  • Incorrect, direct integration can handle this.
  • There is a sine term on the right side of the equation.
  • Incorrect, the sine is not the issue here.
  • The right-hand side contains \(y\text{.}\)
  • Correct! Direct integration is valid only when the right-hand side depends only on the independent variable, in this case \(x\text{.}\)
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(d) Select-the-Best-Answer.
In the differential equation
\begin{equation*} \dfrac{d}{dx}\left[5x \cdot y\right] = \dfrac{1}{x^2}\text{,} \end{equation*}
what is the first step in solving for \(y\text{?}\)
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  • Release \(y\) by integrating both sides with respect to \(x\text{.}\)
  • Correct! Integrating both sides is the first step in solving for \(y\text{.}\)
  • Release \(x\) and \(y\) by integrating both sides with respect to \(y\text{.}\)
  • Incorrect. Integrating both sides with respect to \(y\) would not eliminate the derivative since the derivative is with respect to \(x\text{.}\)
  • Compute the derivative of \(5x \cdot y\) using the product rule.
  • Incorrect. This would actually make the equation more complicated.
  • Isolate \(x\text{.}\)
  • Incorrect. This would not help solve for \(y\text{.}\)
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3. Short-Answer Questions.
(a)
Attempt to apply direct integration to the differential equation
\begin{equation*} \dfrac{dy}{dx} = x + y\text{.} \end{equation*}
Get to the point where it becomes clear that you cannot solve for \(y\) directly. What is the obstacle?
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Solution 1. (a)
  1. 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.
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  2. 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.
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  3. 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, so \(x = 4\) is not a solution to the equation.
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  4. 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, its a bit more tricky. We have to be sure that the functions are equal for all values of \(x\) (or 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.
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    \begin{equation*} x = 0: \end{equation*}
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    \begin{align*} e^0 + \sin 0 + \cos 0 \amp = 1 + 1 \\ 2 \amp = 2 \quad✅ \end{align*}
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    \begin{equation*} x = \dfrac{\pi}{2}: \end{equation*}
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    \begin{align*} e^{\pi/2} + \sin \dfrac{\pi}{2} + \cos \dfrac{\pi}{2} \amp = e^{\pi/2} + 1 \\ e^{\pi/2} + 1 \amp = e^{\pi/2} + 1 \quad✅ \end{align*}
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    \begin{equation*} x = \pi: \end{equation*}
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    \begin{align*} e^{\pi} + \sin \pi + \cos \pi \amp = e^{\pi} + 1 \\ e^{\pi} - 1 \ne\amp\ e^{\pi} + 1 \quad❌ \end{align*}
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    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.
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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.
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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.
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Solution 4. (d)
Integrating both sides gives
\begin{align*} \int \dfrac{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.
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Subsection 🏋️‍♂️ Practice Drills

Exercises Exercises

Select the Solutions.
For each differential equation, select the functions that are solutions to that equation.
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1. \(y''-9y = 0\).
Answer.
a. No
b. Yes
c. Yes
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d. No
e. No
f. No
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g. Yes
h. Yes
i. Yes
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2. \(y'' - 10y' + 25y = 0\).
Answer.
a. Yes
b. No
c. Yes
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d. No
e. Yes
f. Yes
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g. No
h. No
i. Yes
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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.
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3. 🕸️ \(\ y(t) = 2e^{-3t}\).
\(y'' - y' - 12y =\)
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Answer.
\(0\)
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4. 🕸️ \(\ y(t) = 3\sin(t^2)\).
\(y' - ty'' =\)
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Answer.
\(12t^{3}\sin\mathopen{}\left(t^{2}\right)\)
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Subsection ✍🏻 Problems

Exercises Exercises

General Solution.
Find the general solution for each of the following differential equations. Combine constants where appropriate.
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2.
\(\dfrac{d}{dx}[y] = e^{2x}\)
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Answer.
\(y = \dfrac{1}{2}e^{2x} + C\)
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3.
\(\dfrac{d}{dx}[xy] = \cos x\)
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Answer.
\(y = \dfrac{\sin(x)}{x} + \dfrac{C}{x}\)
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4.
\(\dfrac{dy}{dt} = \dfrac{1}{t^2} + t\)
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Answer.
\(y = -\dfrac{1}{t} + \dfrac{1}{2}t^2 + C\)
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6.
\(\left[y\sin x\right]^\prime = \cos^2 x\)
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Answer.
\(y = \dfrac{x}{2\sin(x)} + \dfrac{\cos(x)}{2} + \dfrac{C}{\sin(x)}\)
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7.
\(y' = \ln x + x^2\)
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Answer.
\(y = x \ln(x) - x + \dfrac{1}{3}x^3 + C\)
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8.
\(\dfrac{d}{dP}\left[e^P Q\right] = P\)
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Answer.
\(Q = e^{-P}\left(\dfrac{P^2}{2} + C\right)\)
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Particular Solution.
Find the particular solution for each of the following differential equations with the given initial condition.
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9.
\(\dfrac{dy}{dx} = 3x^2 + 2,\ y(1) = 4\)
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Answer.
\(y = x^3 + 2x + 1\)
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10.
\(\dfrac{d}{dt}[y] = \sin t,\ y(0) = 2\)
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Answer.
\(y = -\cos(t) + 3\)
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11.
\(\dfrac{d}{dx}[x y] = e^x,\ y(1) = 0\)
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Answer.
\(y = \dfrac{e^x}{x} - \dfrac{e}{x}\)
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12.
\(\dfrac{dy}{dx} = x^3 - 4,\ y(2) = 1\)
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Answer.
\(y = \dfrac{x^4}{4} - 4x + 5\)
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13.
\(\dfrac{dR}{dh} = \dfrac{1}{h} + h,\ R(1) = 3\)
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Answer.
\(R = \ln|h| + \dfrac{h^2}{2} + \dfrac{5}{2}\)
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14.
\(y' = \cos x + x^2,\ y(0) = -2\)
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Answer.
\(y = \sin(x) + \dfrac{x^3}{3} - 2\)
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15.
\(\dfrac{d}{dx}\left[e^x y\right] = \sin x,\ y(0) = 2\)
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Answer.
\(y = e^{-x} \left(-\cos x + 3 \right)\)
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16.
\(2y' - 4\sin x = 2, \ y(0) = 5\)
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Answer.
\(y = x - 2 \cos x + 7\)
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17.
\(\left[y\tan(x)\right]^{\prime} = \sec^2(x),\ y\left(\dfrac{\pi}{4}\right) = 1\)
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Answer.
\(y = 1\)
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18. 🕸️ Compute the General Solution.
Given the differential equation
\begin{equation*} y'= e^{2t} - 4t \end{equation*}
Find the general solution.
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Press Activate to submit your answer.
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\(y(t) =\)
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Don’t forget the constant of integration. Do not use scripts on the constant (e.g., \(c_2\)).
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Answer.
\(0.5e^{2t}-2t^{2}+C\)
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19. Solve the Equation.
Solve the initial-value problem
\begin{equation*} 2y' - 4\sin x = 2, \quad y(0) = 5 \text{.} \end{equation*}
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Answer.
\(y = x - 2 \cos x + 7\)
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Exercises Preview of a Future Method

At this point, you should be comfortable solving an equation such as
\begin{equation*} \left[x^7 y \right]^{\prime} = e^x\text{.} \end{equation*}
The problem is that most equations do not start in this form. Instead, they start in another form and, after some algebra, are put into this convenient form and solved. The process of rewriting an equation in this way forms the basis of another technique, the integrating factor method. The question we want to answer here is “what type of equations can be written in this form?”
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1. Give the equation that can be rewritten in the form \(\ds\left[x^7 y \right]^{\prime} = e^x\).
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Rewrite and Solve.
For each equation below, complete the following:
  1. Use the product rule to rewrite each differential equation in the form
    \begin{equation*} y^{\prime} + P(x) y = Q(x)\text{.} \end{equation*}
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  2. Solve the equation.
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2.
\(\left[y \cos x \right]^{\prime} = \dfrac{1}{x^2}\)
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3.
\(\left[y x^5 \right]^{\prime} = x\)
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4.
\(\left[x^{-1}y\right]^{\prime} = e^{2x}\)
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