Separable and linear equations
Separate variables and integrate, or build an integrating factor, then fit the constant to an initial condition.
Question 1
Differential Equations practice question 1
Differential Equations practice topics
Question sets mix direct skills with unfamiliar applications so method selection becomes part of the practice.
Second-order models
Solve constant-coefficient equations through the characteristic equation, and read amplitude, period, and damping from the result.
Systems and phase behavior
Find the eigenvalues of a system matrix and classify the equilibrium as a node, spiral, center, or saddle point.
Laplace transforms and series methods
Transform an initial-value problem into an algebraic one, solve for Y(s), and invert it back to a function of t.
Differential Equations practice questions with worked answers
These 8 questions are printed in full on this page, with every step of the arithmetic written out. Cover the options, solve the question on paper first, and only then open the worked answer to compare your method with the one shown.
Question 1: Solve dy/dx = 3y with the starting condition y(0) = 4.
- y = 4e^(3x)
- y = 3e^(4x)
- y = 4e^x + 3
- y = 4 + 3x
Show the worked answer
- Separate the variables: dy/y = 3 dx.
- Integrate both sides: ln|y| = 3x + C, so y = Ae^(3x).
- Use y(0) = 4: at x = 0, e^0 = 1, so A = 4.
- The solution is y = 4e^(3x). Check: dy/dx = 12e^(3x) = 3 x 4e^(3x).
Answer: y = 4e^(3x)
Question 2: A coffee cools by dT/dt = -0.1(T - 20), starting at 90 degrees C in a 20 degree room. What is its temperature after 10 minutes? Use e^(-1) = 0.368.
- About 45.8 degrees C
- About 33.1 degrees C
- About 25.8 degrees C
- About 70.0 degrees C
Show the worked answer
- Newton's cooling law gives T = 20 + (90 - 20)e^(-0.1t), so T = 20 + 70e^(-0.1t).
- At t = 10 the exponent is -0.1 x 10 = -1.
- 70 x 0.368 = 25.76.
- 20 + 25.76 = 45.76, which rounds to about 45.8 degrees C.
Answer: About 45.8 degrees C
Question 3: What is the integrating factor for dy/dx + 2y = e^x?
- e^(2x)
- e^(-2x)
- 2e^x
- e^(x^2)
Show the worked answer
- The equation is already in the form dy/dx + P(x)y = Q(x), with P(x) = 2.
- The integrating factor is e raised to the integral of P(x) dx.
- The integral of 2 dx is 2x.
- So the integrating factor is e^(2x), and multiplying through makes the left side the derivative of y e^(2x).
Answer: e^(2x)
Question 4: Solve the second-order equation y'' - 5y' + 6y = 0.
- y = C1 e^(2x) + C2 e^(3x)
- y = C1 e^(-2x) + C2 e^(-3x)
- y = C1 e^(5x) + C2 e^(6x)
- y = (C1 + C2 x) e^(3x)
Show the worked answer
- Try y = e^(rx); this turns the equation into r^2 - 5r + 6 = 0.
- Factor: (r - 2)(r - 3) = 0, so r = 2 or r = 3.
- Two different real roots give two independent exponential solutions.
- The general solution is y = C1 e^(2x) + C2 e^(3x).
Answer: y = C1 e^(2x) + C2 e^(3x)
Question 5: A spring-mass system follows y'' + 16y = 0. What is the period of the motion?
- pi / 2
- 2 pi
- pi
- 8 pi
Show the worked answer
- The equation y'' + w^2 y = 0 describes simple harmonic motion with angular frequency w.
- Here w^2 = 16, so w = 4.
- The period is 2 pi / w = 2 pi / 4.
- That simplifies to pi/2, about 1.57 time units per full swing.
Answer: pi / 2
Question 6: A linear system has the matrix [[1, 2], [2, 1]]. Find its eigenvalues and classify the equilibrium at the origin.
- Eigenvalues 3 and -1, so the origin is a saddle point
- Eigenvalues 3 and 1, so the origin is an unstable node
- Eigenvalues -3 and -1, so the origin is a stable node
- Eigenvalues 1 and 2, so the origin is a center
Show the worked answer
- The trace is 1 + 1 = 2 and the determinant is 1(1) - 2(2) = 1 - 4 = -3.
- Solve L^2 - 2L - 3 = 0, which factors as (L - 3)(L + 1) = 0.
- The eigenvalues are 3 and -1.
- One eigenvalue is positive and one is negative, so paths come in along one direction and leave along the other: a saddle point.
Answer: Eigenvalues 3 and -1, so the origin is a saddle point
Question 7: What is the Laplace transform of f(t) = e^(4t)?
- 1/(s - 4), for s > 4
- 1/(s + 4), for s > -4
- 4/s, for s > 0
- s/(s - 4), for s > 4
Show the worked answer
- The Laplace transform integrates e^(-st) f(t) from 0 to infinity.
- With f(t) = e^(4t) the integrand becomes e^(-(s - 4)t).
- That integral converges only when s - 4 > 0, and its value is 1/(s - 4).
- So the transform is 1/(s - 4) for s > 4.
Answer: 1/(s - 4), for s > 4
Question 8: Use Laplace transforms on y' + 3y = 0 with y(0) = 5. What are Y(s) and y(t)?
- Y(s) = 5/(s + 3) and y(t) = 5e^(-3t)
- Y(s) = 5/(s - 3) and y(t) = 5e^(3t)
- Y(s) = 3/(s + 5) and y(t) = 3e^(-5t)
- Y(s) = 5/s and y(t) = 5
Show the worked answer
- Transform each term: the transform of y' is sY(s) - y(0), and y(0) = 5.
- The equation becomes sY(s) - 5 + 3Y(s) = 0.
- Group the Y(s) terms: (s + 3)Y(s) = 5, so Y(s) = 5/(s + 3).
- Reading that backwards gives y(t) = 5e^(-3t). Check: y' = -15e^(-3t) = -3y.
Answer: Y(s) = 5/(s + 3) and y(t) = 5e^(-3t)
How the three Differential Equations difficulty levels differ
The difficulty buttons above change what a question asks of you, not just the size of the numbers. Each example below is taken from the question set on this page.
| Level | What it tests | Example question | Time target |
|---|---|---|---|
| Foundation | One skill at a time, with the numbers kept small enough to check in your head. | Solve dy/dx = 3y with the starting condition y(0) = 4. | About 1 minute |
| Exam | The wording of a real test paper: pick the method first, then carry out two or three steps. | A coffee cools by dT/dt = -0.1(T - 20), starting at 90 degrees C in a 20 degree room. What is its temperature after 10 minutes? Use e^(-1) = 0.368. | 2 to 3 minutes |
| Challenge | Two ideas combined, or a result you have to interpret after the calculation ends. | A linear system has the matrix [[1, 2], [2, 1]]. Find its eigenvalues and classify the equilibrium at the origin. | 4 to 5 minutes |
How to practice Differential Equations effectively
Begin without notes and explain your choice before checking. For every miss, identify whether the cause was a definition, setup, calculation, interpretation, or time decision. Re-solve the question from a blank page, then return to the same skill in a mixed set tomorrow.
What your Differential Equations answer review should show
A useful review shows more than the correct option. Compare the method with your first attempt, locate the earliest incorrect decision, and write one rule that would prevent the same error in a new Differential Equations problem.
Move from mixed Differential Equations questions to a complete course
This page targets flexible Differential Equations question practice. When you need a syllabus-aligned sequence with unit selection, use Differential Equations practice by unit and return here later for mixed retrieval.
Questions about Differential Equations practice
When should I change the difficulty?
Move up after you can solve several questions accurately without hints and explain the method. Move down for one short set when errors show that a definition or setup is still uncertain.
How often should I practice?
Short sessions on several days usually build stronger recall than one long session. Revisit missed Differential Equations skills the next day, then mix them with older topics later in the week.
Where can I review the lessons in order?
Use Differential Equations practice by unit for a syllabus-aligned sequence with unit selection, practice, and a complete answer review.
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