This is our AP Calculus AB unit test on Differential Equations. These questions cover separable differential equation and testing solutions to differential equations. The main idea of this portion of the AP exam is that the dy and dx factors can be “separated,” and then used to solve for y using a method of integration.

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Question 1 |

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

$y = \frac{7}{2}x^2$ | |

$y = \frac{7}{2}x^2 + C$ | |

$y = \frac{7}{2}x^3 + C$ | |

$\frac{1}{2}y^2$ $= \frac{7}{2}x^2 + C$ |

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):

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

$y = \dfrac{7}{2} x^2 + C$

$dy = 7x dx$

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 |

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

$y = \arctan(2x) + C$ | |

$y = \frac{1}{2}\arctan(2x) + C$ | |

$y = \ln |x^2 + 4| + C$ | |

$y = 2x \ln |x^2 + 4| + C$ |

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 $\ln |u| + C$.

$y = \ln |x^2 + 4| + C$

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

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 |

$\dfrac{dy}{dx} \cdot \dfrac{1}{x}$ $= 7x^2 + 5$

$y = \frac{7}{4}x^3 + 5x^2 + C$ | |

$y = \frac{7}{4}x^4 + C$ | |

$y = \frac{7}{3}x^2 + 5x + C$ | |

$y = \frac{7}{4}x^4 + \frac{5}{2}x^2 + C$ |

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 $y$:

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

$y = \dfrac{7}{4} x^4 + \dfrac{5}{2} x^2 + C$

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

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 |

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

$y^{\frac{1}{3}} = Cx^{\frac{1}{7}}$ | |

$y^3 = Cx^7$ | |

$y = Cx^{\frac{1}{7}}$ | |

$y^7 = Cx^{\frac{1}{3}}$ |

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:

$y^{\frac{1}{3}} = Cx^{\frac{1}{7}}$

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

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 |

$\dfrac{dy}{dx} = − \dfrac{x}{y}$

$x^2 − y^2 = C$ | |

$x^2 + y^2 = C$ | |

$x^2 + 2y^2 = C$ | |

$x^2 + \ln|y| = C$ |

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!

$ydy = −xdx$

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 |

$\dfrac{dy}{dx} = xy^2$

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

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

$y = −\dfrac{1}{\frac{1}{2} + C}$ | |

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

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.

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

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 |

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

$\ln|y| = \frac{1}{3}x^2 + C$ | |

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

$\ln|y| = \frac{1}{4}x^4 + C$ | |

$\ln|y| = \frac{1}{5}x^5 + C$ |

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.

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

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 |

$\dfrac{dy}{dx} = \dfrac{y}{1 + x^2},$ $\text{ and }$ $y(0) = 1$

$y = \ln |1 + x^2|$ | |

$y = 2\ln |1 + x^2|$ | |

$y = e^{2\arctan(x)}$ | |

$y = e^{\arctan(x)}$ |

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.

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 |

$\dfrac{dy}{dx} = 7y\tan(x)$, $y(0) = 128$

$y = 128 \cdot \sec^7(x)$ | |

$y = 64 \cdot \sec^7(x)$ | |

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

$y = 128 \cdot \cos^7(x)$ |

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 $C$:

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

$y = 128 \cdot \sec^7(x)$

$\dfrac{1}{7y} dy = \tan(x)dx$

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 |

$\dfrac{dy}{dx} = e^{2x}$, $y(0) = 0$

$y = \frac{1}{2}e^{2x} + \frac{1}{2}$ | |

$y = \frac{1}{2}e^{2x} − \frac{1}{2}$ | |

$y = \frac{1}{2}e^{2x} − \frac{1}{3}$ | |

$y = \frac{1}{2}e^x − \frac{1}{2}$ |

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:

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

$dy = e^{2x} dx$

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 |

$y'' − y' − 30y = 0$ $\text{Identify}$ $\text{if}$ $y = e^{−5x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$\text{Yes}$ | |

$\text{No}$ | |

$\text{Insufficient }$ $\text{Information}$ |

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.

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

$0 = 0$

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 |

$y''' + 12y'' + 47y' + 60y = 0$ $\text{Identify}$ $\text{if}$ $y = e^{−6x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$\text{Yes}$ | |

$\text{No}$ |

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.

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

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 |

$y'' − 4y' + 4y = 0$ $\text{Identify}$ $\text{if}$ $y = xe^{2x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$\text{Yes}$ | |

$\text{No}$ |

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.

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

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 |

$x^2y'' + xy' = 0$ $\text{Confirm}$ $\text{that}$ $y = \ln(x)$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$\text{Yes}$ | |

$\text{No}$ |

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.

$y = \ln(x)$

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

$y''= −\dfrac{1}{x^2}$

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 |

$y^{(4)} − y = 0$ $\text{Identify}$ $\text{if}$ $y = \cos(x)$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$\text{Yes}$ | |

$\text{No}$ |

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.

$y = \cos(x)$

$y' = −\sin(x)$

$y'' = −\cos(x)$

$y'''= \sin(x)$

$y^{(4)} = \cos(x)$

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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