Differential Equations Practice Problems with Answers

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:

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^\circC cools in a 20^\circC room by \frac{dT}{dt} = -k(T - 20). After 10 minutes it is 60^\circC. When will it reach 30^\circC?

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

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:

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:

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

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}:

7. 25 years is 2.5 half-lives:

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:

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, the growth and decay lesson, the logistic growth lesson and the Euler's method lesson. To see what a slope field tells you before you solve, try the slope fields lesson.