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Exercises 4.4 Exercises
💡 Conceptual Quiz.
Answer the following questions to test your understanding of direct integration.
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 \(x\text{.}\)
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.
(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 \(\ln(3x+1)\text{.}\)
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.
(c) True-or-False .
Combining constants is a common practice in differential equations.
True.
Combining constants to simplify the general solution is very common.
False.
Combining constants to simplify the general solution is very common.
(d) True-or-False .
Solving a differential equation by direct integration involves computing a derivative.
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.
2. Multiple Choice.
(a) Select-the-Best-Answer .
How could you solve for \(y\) in the equation
\begin{equation*}
\frac12\frac{dy}{dx} - \tan(2x) = x\text{?}
\end{equation*}
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 \(\tan(2x)\text{.}\)
Answer .
Isolate
\(\dfrac{dy}{dx}\) and integrate both sides with respect to
\(x\text{.}\)
(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?
\(\quad y\)
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 \(21x - 3x^2 + 3\text{.}\)
(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*}
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 \(x\text{.}\)
Answer .
The right-hand side contains
\(y\text{.}\)
(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 \(y\text{?}\)
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 \(y\text{.}\)
Answer .
Integrate both sides with respect to
\(x\text{.}\)
3. Short-Answer Questions.
(a)
Attempt to apply direct integration to the differential equation
\begin{equation*}
\frac{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?
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.
🏋️♂️ Select the General Solution.
Each equation is already in the form
\(\dfrac{dy}{dx} = f(x)\text{,}\) so it can be solved by direct integration. Select its general solution;and don’t forget the constant of integration.
4. \(\dfrac{dy}{dx} = 6x^2\) .
\(\dfrac{dy}{dx} = 6x^2\)
\(y = 2x^3 + C\)
Correct! \(\int 6x^2\,dx = 6\cdot\dfrac{x^3}{3} + C = 2x^3 + C\text{.}\)
\(y = 6x^3 + C\)
Check the power rule: \(\int 6x^2\,dx = 6\cdot\dfrac{x^3}{3} = 2x^3\text{,}\) not \(6x^3\text{.}\)
\(y = 2x^3\)
The antiderivative is right, but every general solution needs an arbitrary constant \(+\,C\text{.}\)
\(y = 12x + C\)
That is the derivative of \(6x^2\text{.}\) Direct integration asks for the antiderivative.
5. \(\dfrac{dy}{dx} = e^{4x}\) .
\(\dfrac{dy}{dx} = e^{4x}\)
\(y = \dfrac{1}{4}e^{4x} + C\)
Correct! \(\int e^{4x}\,dx = \dfrac{1}{4}e^{4x} + C\text{.}\)
\(y = e^{4x} + C\)
Almost; differentiating \(e^{4x}\) brings down a factor of \(4\text{,}\) so integrating divides by \(4\text{.}\)
\(y = 4e^{4x} + C\)
Integrating \(e^{4x}\) divides by \(4\text{;}\) it does not multiply by \(4\text{.}\)
\(y = \dfrac{1}{4}e^{4x}\)
Right form, but don’t forget the \(+\,C\text{.}\)
6. \(\dfrac{dy}{dx} = \sin(3x)\) .
\(\dfrac{dy}{dx} = \sin(3x)\)
\(y = -\dfrac{1}{3}\cos(3x) + C\)
Correct! \(\int \sin(3x)\,dx = -\dfrac{1}{3}\cos(3x) + C\text{.}\)
\(y = \dfrac{1}{3}\cos(3x) + C\)
Sign slip: \(\int \sin u\,du = -\cos u\text{,}\) so the answer is \(-\dfrac{1}{3}\cos(3x) + C\text{.}\)
\(y = -3\cos(3x) + C\)
The inner factor of \(3\) means you divide by \(3\text{,}\) not multiply.
\(y = -\dfrac{1}{3}\cos(3x)\)
Right form, but include the \(+\,C\text{.}\)
7. \(\dfrac{dy}{dx} = \dfrac{1}{x}\) .
\(\dfrac{dy}{dx} = \dfrac{1}{x}\)
\(y = \ln|x| + C\)
Correct! \(\int \dfrac{1}{x}\,dx = \ln|x| + C\text{.}\)
\(y = -\dfrac{1}{x^2} + C\)
The antiderivative of \(\frac{1}{x}\) is a logarithm, not a power;the power rule does not apply when the exponent is \(-1\text{.}\)
\(y = x\ln|x| + C\)
That is not the antiderivative of \(\frac{1}{x}\text{;}\) \(\int \frac{1}{x}\,dx = \ln|x| + C\text{.}\)
\(y = \ln|x|\)
Right function, but don’t drop the \(+\,C\text{.}\)
🏋️♂️ Integrate a Completed Derivative.
In each equation, the left side is already written as a single derivative. Integrate both sides, then solve for
\(y\text{.}\)
8.
\(\dfrac{d}{dx}\left[x^2 y\right] = 8x^3\)
Answer .
\(y = 2x^2 + \dfrac{C}{x^2}\)
9.
\(\dfrac{d}{dx}\left[e^x y\right] = e^x\)
✍🏻 General Solution.
Find the general solution for each of the following differential equations. Combine constants where appropriate.
10.
\(\dfrac{dy}{dx} = 2x - 5\)
11.
\(\dfrac{d}{dx}[y] = e^{2x}\)
Answer .
\(y = \dfrac{1}{2}e^{2x} + C\)
12.
\(\dfrac{d}{dx}[xy] = \cos x\)
Answer .
\(y = \dfrac{\sin(x)}{x} + \dfrac{C}{x}\)
13.
\(\dfrac{dy}{dt} = \dfrac{1}{t^2} + t\)
Answer .
\(y = -\dfrac{1}{t} + \dfrac{1}{2}t^2 + C\)
14.
\(\dfrac{dy}{dx} = x e^x\)
15.
\(\left[y\sin x\right]^\prime = \cos^2 x\)
Answer .
\(y = \dfrac{x}{2\sin(x)} + \dfrac{\cos(x)}{2} + \dfrac{C}{\sin(x)}\)
16.
Answer .
\(y = x \ln(x) - x + \dfrac{1}{3}x^3 + C\)
17.
\(\dfrac{d}{dP}\left[e^P Q\right] = P\)
Answer .
\(Q = e^{-P}\left(\dfrac{P^2}{2} + C\right)\)
✍🏻 Particular Solution.
Find the particular solution for each of the following differential equations with the given initial condition.
18.
\(\dfrac{dy}{dx} = 3x^2 + 2,\ y(1) = 4\)
19.
\(\dfrac{d}{dt}[y] = \sin t,\ y(0) = 2\)
20.
\(\dfrac{d}{dx}[x y] = e^x,\ y(1) = 0\)
Answer .
\(y = \dfrac{e^x}{x} - \dfrac{e}{x}\)
21.
\(\dfrac{dy}{dx} = x^3 - 4,\ y(2) = 1\)
Answer .
\(y = \dfrac{x^4}{4} - 4x + 5\)
22.
\(\dfrac{dR}{dh} = \dfrac{1}{h} + h,\ R(1) = 3\)
Answer .
\(R = \ln|h| + \dfrac{h^2}{2} + \dfrac{5}{2}\)
23.
\(y' = \cos x + x^2,\ y(0) = -2\)
Answer .
\(y = \sin(x) + \dfrac{x^3}{3} - 2\)
24.
\(\dfrac{d}{dx}\left[e^x y\right] = \sin x,\ y(0) = 2\)
Answer .
\(y = e^{-x} \left(-\cos x + 3 \right)\)
25.
\(2y' - 4\sin x = 2, \ y(0) = 5\)
26.
\(\left[y\tan(x)\right]^{\prime} = \sec^2(x),\ y\left(\dfrac{\pi}{4}\right) = 1\)
27. 🕸️ Compute the General Solution.
Given the differential equation
\begin{equation*}
y'= e^{2t} - 4t
\end{equation*}
Find the general solution.
Press Activate to submit your answer.
28. Solve the Equation.
Solve the initial-value problem
\begin{equation*}
2y' - 4\sin x = 2, \quad y(0) = 5 \text{.}
\end{equation*}
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?”
29. Give the equation that can be rewritten in the form \(\ds\left[x^7 y \right]^{\prime} = e^x\) .
Rewrite and Solve.
For each equation below, complete the following:
Use the product rule to rewrite each differential equation in the form
\begin{equation*}
y^{\prime} + P(x) y = Q(x)\text{.}
\end{equation*}
30.
\(\left[y \cos x \right]^{\prime} = \dfrac{1}{x^2}\)
Answer .
\(\ds y = \left(c - \frac{1}{x}\right)\sec x\)
31.
\(\left[y x^5 \right]^{\prime} = x\)
Answer .
\(\ds y = \frac{1}{2x^3} + \frac{c}{x^5}\)
32.
\(\left[x^{-1}y\right]^{\prime} = e^{2x}\)
Answer .
\(\ds y = \frac{x}{2}e^{2x} + cx\)
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