AP Calculus AB — Differential Equations

This is our AP Calculus AB unit test on differential equations. It covers separable differential equations and verifying solutions.

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Question 1
$\dfrac{dy}{dx} = 7x$

A
$y = \frac{7}{2}x^2$
B
$y = \frac{7}{2}x^2 + C$
C
$y = \frac{7}{2}x^3 + C$
D
$\frac{1}{2}y^2$ $= \frac{7}{2}x^2 + C$
Question 1 Explanation: 
The correct answer is (B). This is a separable differential equation. Multiplying both sides by $dx$:

$dy = 7x dx$

To solve for $y$, we must take the integral of both sides (don’t forget the constant of integration +C):

$\displaystyle\int dy = \displaystyle\int (7x)dx$

$y = \dfrac{7}{2} x^2 + C$
Question 2
$\dfrac{dy}{dx}$ $= \dfrac{2x}{4 + x^2}$

A
$y = \arctan(2x) + C$
B
$y = \frac{1}{2}\arctan(2x) + C$
C
$y = \ln |x^2 + 4| + C$
D
$y = 2x \ln |x^2 + 4| + C$
Question 2 Explanation: 
The correct answer is (C). This is a separable differential equation. Moving the dx to the right side gives:

$dy = \dfrac{2x}{(4 + x^2)} dx$

Taking the integral of both sides:

$\displaystyle\int dy = \displaystyle\int \dfrac{2x}{(x^2 + 4)} dx$

Note that the right-hand integral is in the form $\frac{u'}{u'}$, so it can be expressed as $\ln ⁡|u| + C$.

$y = \ln ⁡|x^2 + 4| + C$
Question 3
$\dfrac{dy}{dx} \cdot \dfrac{1}{x}$ $= 7x^2 + 5$

A
$y = \frac{7}{4}x^3 + 5x^2 + C$
B
$y = \frac{7}{4}x^4 + C$
C
$y = \frac{7}{3}x^2 + 5x + C$
D
$y = \frac{7}{4}x^4 + \frac{5}{2}x^2 + C$
Question 3 Explanation: 
The correct answer is (D). Start by multiplying both sides of the equation by $x$. This gives:

$\dfrac{dy}{dx} = 7x^3 + 5x$

Separating the differential equation gives:

$dy = (7x^3 + 5x)dx$

Integrating both sides to solve for $y$:

$\displaystyle\int dy = \displaystyle\int (7x^3 + 5x)dx$

$y = \dfrac{7}{4} x^4 + \dfrac{5}{2} x^2 + C$
Question 4
$\dfrac{dy}{dx} = \dfrac{3y}{7x}$

A
$y^{\frac{1}{3}} = Cx^{\frac{1}{7}}$
B
$y^3 = Cx^7$
C
$y = Cx^{\frac{1}{7}}$
D
$y^7 = Cx^{\frac{1}{3}}$
Question 4 Explanation: 
The correct answer is (A). Start by moving all the $y$ terms to the left side and all the $x$ terms to the right side:

$\dfrac{dy}{3y} = \dfrac{dx}{7x}$

Integrating both sides to solve for $y$:

$\displaystyle\int \dfrac{1}{3y} dy =\displaystyle\int \dfrac{1}{7x} dx$

$\dfrac{1}{3} \ln⁡|y| = \dfrac{1}{7} \ln⁡|x| + C$

Using the exponent log rule, we can rewrite the equation above as:

$\ln ⁡|y^{\frac{1}{3}}|$ $= \ln⁡|x^{\frac{1}{7}}| + C$

Exponentiating both sides by $e$ to get rid of the logarithms:

$y^{\frac{1}{3}}$ $= e^C\cdot x^{\frac{1}{7}}$

Since $e^C$ is still considered an unknown constant, the equation above is equivalent to:

$y^{\frac{1}{3}} = Cx^{\frac{1}{7}}$
Question 5
$\dfrac{dy}{dx} = − \dfrac{x}{y}$

A
$x^2 − y^2 = C$
B
$x^2 + y^2 = C$
C
$x^2 + 2y^2 = C$
D
$x^2 + \ln|y| = C$
Question 5 Explanation: 
The correct answer is (B). Start by moving the $y$ terms to the left side and the $x$ terms to the right side:

$ydy = −xdx$

Integrating both sides:

$\displaystyle\int ydy = \displaystyle\int −xdx$

$\dfrac{1}{2} y^2 = −\dfrac{1}{2} x^2 + C$

$\dfrac{1}{2} y^2 + \dfrac{1}{2} x^2 = C$

We can multiply through by $2$, since $C$ is an arbitrary constant.

$y^2 + x^2 = C$

Fun Fact: This is the equation of a circle!
Question 6
$\dfrac{dy}{dx} = xy^2$

A
$y = −\dfrac{1}{\frac{1}{2}x^2 + C}$
B
$y = −\dfrac{1}{\frac{1}{2}x + C}$
C
$y = −\dfrac{1}{\frac{1}{2} + C}$
D
$y = −\dfrac{1}{\frac{1}{2}x^2} + C$
Question 6 Explanation: 
The correct answer is (A). Begin by separating the $y$ and $x$ variables to different sides of the equation:

$\dfrac{1}{y^2}dy = xdx$

Integrating both sides:

$\displaystyle\int \dfrac{1}{y^2}dy = \displaystyle\int xdx$

$ −\dfrac{1}{y} = \dfrac{1}{2}x^2 + C$

Rearranging and solving for $y$:

$y = −\dfrac{1}{\frac{1}{2} x^2 + C}$

The $C$ must be divided in the fraction as well while rearranging.
Question 7
$\dfrac{dy}{dx} = x^2y$

A
$\ln|y| = \frac{1}{3}x^2 + C$
B
$\ln|y| = \frac{1}{3}x^3 + C$
C
$\ln|y| = \frac{1}{4}x^4 + C$
D
$\ln|y| = \frac{1}{5}x^5 + C$
Question 7 Explanation: 
The correct answer is (B). Start by separating the variables:

$\dfrac{1}{y} dy = x^2 dx$

Integrate both sides to solve for $y$:

$\ln⁡|y| = \dfrac{1}{3} x^3 + C$

We could exponentiate both sides, but the given answer choices are not simplified.
Question 8
$\dfrac{dy}{dx} = \dfrac{y}{1 + x^2},$ $\text{ and }$ $y(0) = 1$

A
$y = \ln |1 + x^2|$
B
$y = 2\ln |1 + x^2|$
C
$e^{2\arctan(x)}$
D
$e^{\arctan(x)}$
Question 8 Explanation: 
The correct answer is (D).
Begin by separating the variables.

$\dfrac{1}{y}dy = \dfrac{1}{1 + x^2}dx$

Integrating both sides gives the equation below.

$\ln⁡(y) = \arctan⁡(x) + C$

Exponentiating both sides gives the equation below.

$y = Ce^{\arctan⁡(x)}$

Using the initial condition given allows us to solve for the constant of integration.

$1 = C \cdot e^{\arctan⁡(0)}$

$C = 1$

Substitute this value back into the function.

$y = e^{\arctan⁡(x)}$

The process of solving for the constant of integration using an initial condition is very simple but should only be done after the differential equation is solved.
Question 9
$\dfrac{dy}{dx} = 7y\tan(x)$, $y(0) = 128$

A
$y = 128 \cdot \sec^7(x)$
B
$y = 64 \cdot \sec^7(x)$
C
$y = 128^{\frac{1}{7}} \cdot \sec^7(x)$
D
$y = 128 \cdot \cos^7(x)$
Question 9 Explanation: 
The correct answer is (A). Begin by separating the variables:

$\dfrac{1}{7y} dy = \tan⁡(x)dx$

Integrate both sides to solve for $y$.

$\dfrac{1}{7} \ln⁡|y| = −\ln ⁡|\cos⁡(x)| + C$

Exponentiate both sides by power $e$.

$y^{\frac{1}{7}} = C \cdot \sec⁡(x)$

Plugging in the initial value for $y(0)$:

$128^{\frac{1}{7}} = C \cdot \sec⁡(0)$

$C = 128^{\frac{1}{7}}$

Using the new value for $C$:

$y^{\frac{1}{7}}128^{\frac{1}{7}} \cdot \sec⁡(x)$

$y = 128 \cdot \sec^7⁡(x)$
Question 10
$\dfrac{dy}{dx} = e^{2x}$, $y(0) = 0$

A
$y = \frac{1}{2}e^{2x} + \frac{1}{2}$
B
$y = \frac{1}{2}e^{2x} − \frac{1}{2}$
C
$y = \frac{1}{2}e^{2x} − \frac{1}{3}$
D
$y = \frac{1}{2}e^x − \frac{1}{2}$
Question 10 Explanation: 
The correct answer is (B). Begin by separating variables:

$dy = e^{2x} dx$

Integrating both sides gives:

$y = \dfrac{1}{2} e^{2x} + C$

Using the initial value given:

$0 = \dfrac{1}{2} e^0 + C$

$C = −\dfrac{1}{2}$

Substituting this value of $C$ back into the solution:

$y = \dfrac{1}{2} e^{2x} −\dfrac{1}{2}$
Question 11
$y'' − y' − 30y = 0$

$\text{Identify}$ $\text{if}$ $y = e^{−5x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

A
$\text{Yes}$
B
$\text{No}$
C
$\text{Insufficient }$ $\text{Information}$
Question 11 Explanation: 
The correct answer is (A). The first and second derivatives of $y$ are $−5e^{−5x}$ and $25e^{−5x}$ respectively. Plugging in the values for $y$, $y'$ and $y''$ into the given differential equation gives:

$(25e^{−5x}− ( −5e^{−5x} ) − 30(e^{−5x} )$ $= 0$

$0 = 0$

Since the left side is equal to $0$, the equation is true. This means $y=e^{−5x}$ is a valid solution to the differential equation.
Question 12
$y''' + 12y'' + 47y' + 60y = 0$

$\text{Identify}$ $\text{if}$ $y = e^{−6x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

A
$\text{Yes}$
B
$\text{No}$
Question 12 Explanation: 
The correct answer is (B). Taking the first three derivatives of $y$:

$y = e^{−6x}$, $\, y' = −6e^{−6x}$, $\, y'' = 36e^{−6x}$, $\, y''' = −216e^{−6x}$

Plugging in the values above into the differential equation:

$(−216e^{−6x} ) + 12(36e^{−6x} )$ $+ ~47(−6e^{−6x} ) + 60(e^{−6x})$

$= e^{−6x} (−216 + 12 \cdot 36 + 47 \cdot −6 + 60)$ $= −6e^{−6x} ≠ 0$

Since the result is not zero, $e^{−6x}$ is not a valid solution for the differential equation.
Question 13
$y'' − 4y' + 4y = 0$

$\text{Identify}$ $\text{if}$ $y = xe^{2x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

A
$\text{Yes}$
B
$\text{No}$
Question 13 Explanation: 
The correct answer is (A). Taking the first two derivatives of $y$ using the product rule:

$y = xe^{2x}$, $\, y' = e^{2x} + 2xe^{2x}$, $\, y'' = 4e^{2x} + 4xe^{2x}$

Plugging these values back into the differential equation:

$4e^{2x} + 4xe^{2x} − 4(e^{2x} + 2xe^{2x} ) + 4(xe^{2x} )$

$= 4e^{2x} + 4xe^{2x} − 4e^{2x} − 8xe^{2x} + 4xe^{2x}$ $= 0$

Since the result is $0$, $y = xe^{2x}$ is a valid solution for the differential equation.
Question 14
$x^2y'' + xy' = 0$

$\text{Confirm}$ $\text{that}$ $y = \ln(x)$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

A
$\text{Yes}$
B
$\text{No}$
Question 14 Explanation: 
The correct answer is (A). The derivatives of the solution are as follows.

$y = \ln⁡(x)$

$y' = \dfrac{1}{x}$

$y''= −\dfrac{1}{x^2}$

Substituting these values back into the differential equation gives the following.

$x^2 \cdot −\dfrac{1}{x^2} + x \cdot \dfrac{1}{x} = 0$

$−1 + 1 = 0$

$0 = 0$

This confirms that the given solution is correct.
Question 15
$y^{(4)} − y = 0$

$\text{Identify}$ $\text{if}$ $y = \cos(x)$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

A
$\text{Yes}$
B
$\text{No}$
Question 15 Explanation: 
The correct answer is (A). Taking the first $4$ derivatives of $y$:

$y = \cos⁡(x)$
$y' = −\sin⁡(x)$
$y'' = −\cos⁡(x)$
$y'''= \sin⁡(x)$
$y^{(4)} = \cos⁡(x)$

Substituting into the differential equation:

$\cos⁡(x) − \cos⁡(x) = 0$ Since the result is $0$, we see that $y = \cos⁡(x)$ is a valid solution for the differential equation.
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