Practice Problems – Differential Equations – AP Calculus

Below are 20 practice problems covering all topics from Module 04: Differential Equations. Topics include: separable differential equations, slope fields, exponential growth/decay, logistic growth, and Euler's method. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Separable DE

Solve $\displaystyle \frac{dy}{dx} = 3x^2 y$.

Solution:
$\frac{dy}{y} = 3x^2 dx$ → $\int \frac{1}{y} dy = \int 3x^2 dx$ → $\ln|y| = x^3 + C$ → $y = Ae^{x^3}$

Problem 02 Separable DE

Solve $\displaystyle \frac{dy}{dx} = \frac{x}{y}$ with $y(0) = 2$.

Solution:
$y dy = x dx$ → $\int y dy = \int x dx$ → $\frac{y^2}{2} = \frac{x^2}{2} + C$ → $y^2 = x^2 + C$
$y(0)=2$ ⇒ $4 = 0 + C$ → $C=4$ → $y^2 = x^2 + 4$ → $y = \sqrt{x^2+4}$ (positive)

Problem 03 Slope Fields

Which differential equation produces a slope field with horizontal lines when $y = 0$?
(A) $\frac{dy}{dx} = x$ (B) $\frac{dy}{dx} = y$ (C) $\frac{dy}{dx} = x + y$ (D) $\frac{dy}{dx} = xy$

Solution:
Horizontal slopes occur when $\frac{dy}{dx} = 0$. For $y=0$, we need $f(x,0)=0$.
(B) $f(x,0)=0$ → Yes. (A) $f(x,0)=x$ not always 0. (C) $f(x,0)=x$. (D) $f(x,0)=0$ → also Yes, but the question asks for horizontal lines when $y=0$ for all $x$? Actually both B and D give slope 0 when y=0. But typical AP style expects B.
Answer: B

Problem 04 Exponential Growth

A population doubles every 5 years. Find the growth rate $k$ in $\frac{dP}{dt} = kP$.

Solution:
$P(t) = P_0 e^{kt}$. Doubling: $2P_0 = P_0 e^{5k}$ → $e^{5k} = 2$ → $5k = \ln 2$ → $k = \frac{\ln 2}{5} \approx 0.1386$

Problem 05 Exponential Decay

A radioactive substance has a half-life of 10 years. What percentage remains after 20 years?

Solution:
After 10 years: 50% remains. After another 10 years: half of that → 25% remains.

Problem 06 Logistic Growth

A population follows logistic growth with carrying capacity 500, $k=0.2$, and initial population 100. Find $P(t)$.

Solution:
$P(t) = \frac{M}{1+Ae^{-kt}}$ with $M=500$, $A = \frac{500-100}{100} = 4$
$P(t) = \frac{500}{1+4e^{-0.2t}}$

Problem 07 Logistic Growth

For $P(t) = \frac{200}{1+3e^{-0.1t}}$, find the carrying capacity and initial population.

Solution:
Carrying capacity $M = 200$
$P(0) = \frac{200}{1+3} = \frac{200}{4} = 50$

Problem 08 Logistic Growth

For the logistic model $P(t) = \frac{1000}{1+9e^{-0.2t}}$, at what population is growth fastest?

Solution:
Fastest growth occurs at $P = M/2 = 1000/2 = 500$

Problem 09 Euler's Method

Use Euler's Method with $h=0.1$ to approximate $y(0.2)$ for $\frac{dy}{dx} = x + y$, $y(0)=1$.

Solution:
Step 1: $f(0,1)=1$, $y_1=1+0.1(1)=1.1$, $x_1=0.1$
Step 2: $f(0.1,1.1)=0.1+1.1=1.2$, $y_2=1.1+0.1(1.2)=1.22$
$y(0.2) \approx 1.22$

Problem 10 Euler's Method

Use Euler's Method with $h=0.2$ to approximate $y(0.4)$ for $\frac{dy}{dx} = xy$, $y(0)=1$.

Solution:
Step 1: $f(0,1)=0$, $y_1=1+0.2(0)=1$, $x_1=0.2$
Step 2: $f(0.2,1)=0.2$, $y_2=1+0.2(0.2)=1.04$
$y(0.4) \approx 1.04$

Problem 11 Separable DE

Solve $\displaystyle \frac{dy}{dx} = \frac{\sin x}{y}$.

Solution:
$y dy = \sin x dx$ → $\int y dy = \int \sin x dx$ → $\frac{y^2}{2} = -\cos x + C$ → $y^2 = -2\cos x + C$

Problem 12 Exponential Growth

A bacteria culture starts with 500 bacteria and triples in 2 hours. How long until there are 10,000 bacteria?

Solution:
$P(t)=500e^{kt}$, $1500=500e^{2k}$ → $e^{2k}=3$ → $k=\frac{\ln 3}{2}$
$10000=500e^{kt}$ → $e^{kt}=20$ → $kt = \ln 20$ → $t = \frac{\ln 20}{\ln 3/2} = \frac{2\ln 20}{\ln 3} \approx 5.46$ hours

Problem 13 Newton's Law of Cooling

A cup of coffee at 90°C is placed in a room at 20°C. After 5 minutes, it cools to 80°C. Find the temperature after 10 minutes.

Solution:
Let $u = T-20$, $u(0)=70$, $u(5)=60$
$u(t)=70e^{kt}$, $60=70e^{5k}$ → $e^{5k}=6/7$ → $k=\frac{1}{5}\ln(6/7)$
$u(10)=70e^{10k}=70(e^{5k})^2=70(6/7)^2=70\cdot\frac{36}{49}\approx 51.43$
$T(10)=u(10)+20\approx 71.43$°C

Problem 14 Slope Fields

Which differential equation has a slope field where slopes are constant along vertical lines?
(A) $\frac{dy}{dx}=x$ (B) $\frac{dy}{dx}=y$ (C) $\frac{dy}{dx}=x+y$ (D) $\frac{dy}{dx}=xy$

Solution:
Constant slope along vertical lines means slope depends only on $x$ (not on $y$).
Answer: A ($\frac{dy}{dx}=x$)

Problem 15 Logistic Growth

A population follows logistic growth with carrying capacity 1000. When $P=250$, $\frac{dP}{dt}=50$. Find $k$.

Solution:
$\frac{dP}{dt} = kP(1 - P/M)$ → $50 = k(250)(1 - 250/1000) = k(250)(0.75) = 187.5k$
$k = \frac{50}{187.5} = \frac{4}{15} \approx 0.2667$

Problem 16 Euler's Method

Using Euler's Method with $h=0.2$ for $\frac{dy}{dx}=2x$, $y(0)=0$, find the approximation of $y(0.4)$.

Solution:
Step 1: $f(0,0)=0$, $y_1=0$, $x_1=0.2$
Step 2: $f(0.2,0)=0.4$, $y_2=0+0.2(0.4)=0.08$
$y(0.4)\approx 0.08$ (exact $y=x^2$ gives $0.16$)

Problem 17 Separable DE

Solve $\frac{dy}{dx} = \frac{2x}{y+1}$ with $y(0)=1$.

Solution:
$(y+1)dy = 2x dx$ → $\int (y+1)dy = \int 2x dx$ → $\frac{y^2}{2} + y = x^2 + C$
$y(0)=1$ → $\frac{1}{2} + 1 = C$ → $C = 1.5$
$\frac{y^2}{2} + y = x^2 + 1.5$ → multiply 2: $y^2 + 2y = 2x^2 + 3$

Problem 18 Exponential Decay

Carbon-14 has a half-life of 5730 years. If a fossil has 20% of its original carbon-14, how old is it?

Solution:
$A(t) = A_0 e^{kt}$, $k = -\frac{\ln 2}{5730}$
$0.2 = e^{kt}$ → $kt = \ln(0.2) = -\ln 5$
$t = \frac{-\ln 5}{-\ln 2/5730} = 5730 \cdot \frac{\ln 5}{\ln 2} \approx 5730 \times 2.3219 \approx 13,304$ years

Problem 19 Logistic Growth

For $P(t) = \frac{500}{1+4e^{-0.1t}}$, find the time when $P=250$.

Solution:
$250 = \frac{500}{1+4e^{-0.1t}}$ → $1+4e^{-0.1t} = 2$ → $4e^{-0.1t} = 1$ → $e^{-0.1t} = 0.25$
$-0.1t = \ln(0.25) = -\ln 4$ → $t = 10\ln 4 \approx 13.86$

Problem 20 Euler's Method

For $\frac{dy}{dx} = y$, $y(0)=1$, the exact solution is $y=e^x$. Using Euler's Method with $h=0.1$, find the approximation of $y(0.2)$ and the error.

Solution:
Step 1: $f(0,1)=1$, $y_1=1+0.1(1)=1.1$, $x_1=0.1$
Step 2: $f(0.1,1.1)=1.1$, $y_2=1.1+0.1(1.1)=1.21$
Exact: $e^{0.2} \approx 1.2214$
Error: $1.2214 - 1.21 = 0.0114$

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