# Differential Equations Practice Problems with Answers

> 12 differential equation practice problems with answers: separable equations, exponential growth and decay, cooling, logistic growth and Euler's method.

Source: https://practicecalculus.com/practice-problems/differential-equations
Updated: 2026-11-06

$$ \frac{dy}{dx} = g(x)\,h(y) \;\Rightarrow\; \int\frac{dy}{h(y)} = \int g(x)\,dx $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2 and AP Calculus, Unit 7
- **You need:** [Antiderivatives](/practice-problems/antiderivatives), logs and exponentials
- **Main idea:** Separate the variables, integrate both sides, then use the initial condition to find $C$
- **Practice with new numbers:** [Separable equations lesson](/learn/calculus-2/improper-integrals-and-applications/separable-equations)

## Key takeaways

- Separate: move every $y$ to the side with $dy$ and every $x$ to the side with $dx$, then integrate both sides.
- Add the constant right after integrating, before you solve for $y$.
- The equation $\frac{dy}{dt} = ky$ always gives $y = y_0e^{kt}$. Learn to spot it in word problems.

These **differential equations practice problems** cover separable equations, exponential growth and decay, Newton's law of cooling, logistic growth, equilibrium solutions and Euler's method. Try each one first. The answers come first, then full solutions.

The method for a separable equation:

$$\frac{dy}{dx} = g(x)\,h(y) \quad\Rightarrow\quad \int\frac{1}{h(y)}\,dy = \int g(x)\,dx$$

Separate, integrate both sides, add $C$, then solve for $y$.

## Problems

**Separable equations**

1. $\dfrac{dy}{dx} = 6x^2$, $y(0) = 2$
2. $\dfrac{dy}{dx} = 2xy$, $y(0) = 5$
3. $\dfrac{dy}{dx} = \dfrac{x}{y}$, $y(0) = 3$
4. $\dfrac{dy}{dx} = y^2$, $y(0) = 1$
5. $\dfrac{dy}{dx} = y\cos x$, $y(0) = 2$

**Growth and decay**

6. $\dfrac{dy}{dt} = -0.2y$, $y(0) = 100$. Find $y(5)$.
7. A sample of 80 g has a half-life of 10 years. How much is left after 25 years?
8. A bacteria culture starts at 500 and doubles every 3 hours. When does it reach 8000?
9. Coffee at $90^\circ$C cools in a $20^\circ$C room by $\frac{dT}{dt} = -k(T - 20)$. After 10 minutes it is $60^\circ$C. When will it reach $30^\circ$C?

**Qualitative and numerical**

10. For $\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{1000}\right)$, find the carrying capacity and the population where growth is fastest.
11. Find the equilibrium solutions of $\dfrac{dy}{dx} = y(y - 3)$ and say whether each is stable.
12. Use Euler's method with $h = 0.1$ and two steps to estimate $y(0.2)$ for $\dfrac{dy}{dx} = x + y$, $y(0) = 1$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $y = 2x^3 + 2$ |
| 2 | $y = 5e^{x^2}$ |
| 3 | $y = \sqrt{x^2 + 9}$ |
| 4 | $y = \dfrac{1}{1 - x}$ |
| 5 | $y = 2e^{\sin x}$ |
| 6 | $y(5) = 100e^{-1} \approx 36.8$ |
| 7 | $80\left(\frac{1}{2}\right)^{2.5} \approx 14.1$ g |
| 8 | $12$ hours |
| 9 | $t = \dfrac{10\ln 7}{\ln(7/4)} \approx 34.8$ minutes |
| 10 | Carrying capacity $1000$; fastest at $P = 500$ |
| 11 | $y = 0$ stable, $y = 3$ unstable |
| 12 | $y(0.2) \approx 1.22$ |

## Step-by-step solutions

**1.** No separating needed. Integrate: $y = 2x^3 + C$. Since $y(0) = 2$, $C = 2$.

**2.** Separate: $\frac{dy}{y} = 2x\,dx$. Integrate: $\ln|y| = x^2 + C$. Exponentiate: $y = Ae^{x^2}$. Since $y(0) = 5$, $A = 5$:

$$y = 5e^{x^2}$$

**3.** Separate: $y\,dy = x\,dx$. Integrate: $\frac{y^2}{2} = \frac{x^2}{2} + C$, so $y^2 = x^2 + C'$. With $y(0) = 3$, $C' = 9$. Take the positive root because $y(0) > 0$:

$$y = \sqrt{x^2 + 9}$$

**4.** Separate: $y^{-2}\,dy = dx$. Integrate: $-\frac{1}{y} = x + C$. With $y(0) = 1$, $C = -1$, so $-\frac{1}{y} = x - 1$:

$$y = \frac{1}{1 - x}$$

The solution blows up at $x = 1$, even though the equation looks harmless.

**5.** Separate: $\frac{dy}{y} = \cos x\,dx$. Integrate: $\ln|y| = \sin x + C$, so $y = Ae^{\sin x}$, and $y(0) = 2$ gives $A = 2$.

**6.** This is $\frac{dy}{dt} = ky$ with $k = -0.2$, so $y = 100e^{-0.2t}$:

$$y(5) = 100e^{-1} \approx 36.8$$

**7.** 25 years is $2.5$ half-lives:

$$80\left(\tfrac{1}{2}\right)^{2.5} \approx 80(0.1768) \approx 14.1 \text{ g}$$

**8.** $8000 = 500\cdot 16 = 500\cdot 2^4$. That is 4 doublings of 3 hours each, so 12 hours.

**9.** Let $D = T - 20$. Then $\frac{dD}{dt} = -kD$, so $T = 20 + 70e^{-kt}$. At $t = 10$: $40 = 70e^{-10k}$, so $k = \frac{\ln(7/4)}{10} \approx 0.056$. For $T = 30$: $10 = 70e^{-kt}$, so $e^{-kt} = \frac{1}{7}$ and:

$$t = \frac{\ln 7}{k} = \frac{10\ln 7}{\ln(7/4)} \approx 34.8 \text{ minutes}$$

**10.** The equation has the form $kP\left(1 - \frac{P}{L}\right)$ with $L = 1000$. The right side is a downward parabola in $P$ with roots $0$ and $1000$, so it is largest at the vertex, $P = 500$.

**11.** Set $y(y - 3) = 0$: $y = 0$ and $y = 3$. Between them, $y(y - 3) < 0$, so solutions fall toward $0$. Above $3$, $y(y - 3) > 0$, so solutions rise away from $3$. Below $0$, the slope is positive, so solutions rise toward $0$. That makes $0$ stable and $3$ unstable.

**12.** Use $y_{n+1} = y_n + h\cdot f(x_n, y_n)$:

- Step 1: $y_1 = 1 + 0.1(0 + 1) = 1.1$
- Step 2: $y_2 = 1.1 + 0.1(0.1 + 1.1) = 1.22$

The exact solution is $y = 2e^x - x - 1$, so $y(0.2) \approx 1.2428$. Euler's method underestimates here because the solution curves are concave up.

## Common mistakes on these problems

- **Adding $C$ too late.** In problem 2, $y = e^{x^2} + C$ is wrong. Add $C$ right after integrating.
- **Not separating fully.** $\frac{dy}{dx} = x + y$ cannot be separated. That is why problem 12 uses Euler's method.
- **Losing the units of time.** In problem 7, divide 25 years by the 10-year half-life before using the formula.
- **Using $T$ instead of $T - 20$.** In cooling problems, the exponential describes the difference from room temperature.

## Keep practicing

For unlimited problems with instant feedback, try the [separable equations lesson](/learn/calculus-2/improper-integrals-and-applications/separable-equations), the [growth and decay lesson](/learn/calculus-2/improper-integrals-and-applications/growth-and-decay), the [logistic growth lesson](/learn/calculus-2/improper-integrals-and-applications/logistic-growth) and the [Euler's method lesson](/learn/calculus-2/improper-integrals-and-applications/eulers-method). To see what a slope field tells you before you solve, try the [slope fields lesson](/learn/calculus-2/improper-integrals-and-applications/slope-fields).

## Frequently asked questions

### What is a separable differential equation?

One that can be written as $\frac{dy}{dx} = g(x)h(y)$, a function of $x$ times a function of $y$. Then you can divide by $h(y)$ and integrate each side on its own.

### Why do you add C before solving for y?

Because the constant gets transformed along with everything else. In problem 2, $\ln|y| = x^2 + C$ becomes $y = Ae^{x^2}$, where $A = \pm e^{C}$. Adding $C$ at the end instead would give $e^{x^2} + C$, which is a different and wrong function.

### What does the carrying capacity mean in logistic growth?

It is the population size the environment can support. In $\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$, the population approaches $L$, and it grows fastest when $P = \frac{L}{2}$.