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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.

  • Combining constants to simplify the general solution is very common.

  • False.

  • Combining constants to simplify the general solution is very common.
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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.

  • Since this equation is in the form (4.1), direct integration applies.

  • False.

  • Since this equation is in the form (4.1), direct integration applies.
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Solution.
  1. False. To solve for \(y\text{,}\) we need to integrate both sides with respect to \(x\text{,}\) not differentiate. Differentiating both sides would result in a second derivative on the left side.
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  2. True. Integrating both sides gives \(y = \int \ln(3x+1)\ dx\text{,}\) which is the antiderivative of \(\ln(3x+1)\text{.}\)
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  3. True. Combining constants to simplify the general solution is very common in differential equations.
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  4. False. Direct integration involves integrating both sides of the equation, not computing a derivative.
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  5. True. Since this equation is in the form (4.1), direct integration applies.
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Answer.
  1. False
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  2. True
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  3. True
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  4. False
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  5. True
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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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Solution.
  1. We need to isolate \(\dfrac{dy}{dx}\) first, then integrate both sides with respect to \(x\text{.}\) Isolating gives \(\dfrac{dy}{dx} = 2\tan(2x) + 2x\text{,}\) and then we integrate to find \(y\text{.}\)
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  2. First, isolate \(y'\text{:}\)
    \begin{align*} \frac13 y' \amp = 7x - x^2 + 1 \\ y' \amp = 21x - 3x^2 + 3 \end{align*}
    The solution is the antiderivative of \(21x - 3x^2 + 3\text{.}\)
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  3. Direct integration requires that the right-hand side depends only on the independent variable (in this case, \(x\)). Since \(\sin(x+y)\) contains \(y\text{,}\) direct integration cannot be applied.
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  4. The first step is to integrate both sides with respect to \(x\text{.}\) This will release \(y\) from the derivative, allowing us to solve for it.
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Answer.
  1. Isolate \(\dfrac{dy}{dx}\) and integrate both sides with respect to \(x\text{.}\)
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  2. \(\displaystyle 21x - 3x^2 + 3\)
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  3. The right-hand side contains \(y\text{.}\)
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  4. Integrate both sides with respect to \(x\text{.}\)
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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.
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 \(y\) terms into a single \(y\) on the left-hand side.
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Answer.
The obstacle is that the right-hand side contains \(y\text{,}\) which means we cannot evaluate \(\int y\ dx\) when integrating with respect to \(x\text{.}\) We would need to know \(y\) as a function of \(x\) to proceed, but that is what we are trying to find.
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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. \(y''-9y = 0\).
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2. \(y'' - 10y' + 25y = 0\).
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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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4. 🕸️ \(\ y(t) = 3\sin(t^2)\).
\(y' - ty'' =\)
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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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3.
\(\dfrac{d}{dx}[xy] = \cos x\)
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4.
\(\dfrac{dy}{dt} = \dfrac{1}{t^2} + t\)
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6.
\(\left[y\sin x\right]^\prime = \cos^2 x\)
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8.
\(\dfrac{d}{dP}\left[e^P Q\right] = P\)
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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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10.
\(\dfrac{d}{dt}[y] = \sin t,\ y(0) = 2\)
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11.
\(\dfrac{d}{dx}[x y] = e^x,\ y(1) = 0\)
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12.
\(\dfrac{dy}{dx} = x^3 - 4,\ y(2) = 1\)
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13.
\(\dfrac{dR}{dh} = \dfrac{1}{h} + h,\ R(1) = 3\)
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14.
\(y' = \cos x + x^2,\ y(0) = -2\)
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15.
\(\dfrac{d}{dx}\left[e^x y\right] = \sin x,\ y(0) = 2\)
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16.
\(2y' - 4\sin x = 2, \ y(0) = 5\)
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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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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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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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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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You have attempted of activities on this page.
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