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Differential Equations covers first-order methods, linear equations with constant coefficients, Laplace transforms, systems, and series solutions. Exams are classification-first: recognize the equation type, then execute the matching method cleanly. Worked practice exams train that recognition reflex and catch the algebra slips that cost the most marks.

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Topics typically tested

  • Separable & linear first-order equations
  • Exact equations & integrating factors
  • Second-order linear with constant coefficients
  • Undetermined coefficients & variation of parameters
  • Laplace transform methods
  • Systems of linear ODEs

Sample Differential Equations exam questions

Three original questions in the style of a real Differential Equations exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceSecond-order linear with constant coefficients

Solve the initial value problem

y′′+4y′+13y=0,y(0)=1,y′(0)=−2.y'' + 4y' + 13y = 0, \qquad y(0) = 1, \quad y'(0) = -2.
  1. A.

    y=e−2tcos⁡3ty = e^{-2t}\cos 3t

  2. B.

    y=e2tcos⁡3ty = e^{2t}\cos 3t

  3. C.

    y=e−2t(cos⁡3t−23sin⁡3t)y = e^{-2t}\left(\cos 3t - \tfrac{2}{3}\sin 3t\right)

  4. D.

    y=e−4tcos⁡6ty = e^{-4t}\cos 6t

Worked solution
Answer:A

This is a homogeneous linear equation with constant coefficients, so try y=erty = e^{rt}; the characteristic equation is

r2+4r+13=0⟹r=−4±16−522=−4±6i2=−2±3i.r^2 + 4r + 13 = 0 \quad\Longrightarrow\quad r = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i.

Complex roots α±βi\alpha \pm \beta i give the general solution y=eαt(c1cos⁡βt+c2sin⁡βt)y = e^{\alpha t}(c_1\cos\beta t + c_2\sin\beta t), so here

y(t)=e−2t(c1cos⁡3t+c2sin⁡3t).y(t) = e^{-2t}\left(c_1\cos 3t + c_2\sin 3t\right).

Apply y(0)=1y(0) = 1: e0(c1+0)=1e^{0}(c_1 + 0) = 1, so c1=1c_1 = 1.

For y′(0)y'(0), differentiate with the product rule, keeping the derivative of the exponential factor:

y′(t)=−2e−2t(c1cos⁡3t+c2sin⁡3t)+e−2t(−3c1sin⁡3t+3c2cos⁡3t).y'(t) = -2e^{-2t}\left(c_1\cos 3t + c_2\sin 3t\right) + e^{-2t}\left(-3c_1\sin 3t + 3c_2\cos 3t\right).

At t=0t = 0: y′(0)=−2c1+3c2=−2+3c2y'(0) = -2c_1 + 3c_2 = -2 + 3c_2. Setting this equal to −2-2 gives c2=0c_2 = 0.

So y=e−2tcos⁡3ty = e^{-2t}\cos 3t. Substituting back: y′=e−2t(−2cos⁡3t−3sin⁡3t)y' = e^{-2t}(-2\cos 3t - 3\sin 3t) and y′′=e−2t(−5cos⁡3t+12sin⁡3t)y'' = e^{-2t}(-5\cos 3t + 12\sin 3t), so y′′+4y′+13y=e−2t[(−5−8+13)cos⁡3t+(12−12)sin⁡3t]=0y'' + 4y' + 13y = e^{-2t}\big[(-5 - 8 + 13)\cos 3t + (12 - 12)\sin 3t\big] = 0, as required.

The tempting wrong answer is C: it comes from differentiating only the trigonometric factor and forgetting the −2e−2t-2e^{-2t} term from the product rule, which leads to 3c2=−23c_2 = -2. Option B has the wrong sign on the real part of the roots (the solution would grow, not decay), and option D forgets to divide by 2 in the quadratic formula.

y(t)=e−2tcos⁡3ty(t) = e^{-2t}\cos 3t (option A).

Question 2Short answerSeparable & linear first-order equations

Solve the initial value problem

t dydt+2y=t3,y(1)=1,t>0.t\,\frac{dy}{dt} + 2y = t^3, \qquad y(1) = 1, \quad t > 0.
Worked solution
Answer:

y=t35+45t2y = \dfrac{t^3}{5} + \dfrac{4}{5t^2}

The equation is linear in yy but not in standard form. Divide by tt (allowed since t>0t > 0):

dydt+2t y=t2.\frac{dy}{dt} + \frac{2}{t}\,y = t^2.

For y′+p(t)y=q(t)y' + p(t)y = q(t) the integrating factor is μ(t)=e∫p dt\mu(t) = e^{\int p\,dt}; here

μ(t)=e∫2/t dt=e2ln⁡t=t2.\mu(t) = e^{\int 2/t\,dt} = e^{2\ln t} = t^2.

Multiplying the standard form by μ\mu makes the left side an exact derivative, which is the whole point of the method:

t2y′+2t y=t4⟹ddt(t2y)=t4.t^2 y' + 2t\,y = t^4 \quad\Longrightarrow\quad \frac{d}{dt}\left(t^2 y\right) = t^4.

Integrate both sides:

t2y=t55+C⟹y=t35+Ct2.t^2 y = \frac{t^5}{5} + C \quad\Longrightarrow\quad y = \frac{t^3}{5} + \frac{C}{t^2}.

Apply y(1)=1y(1) = 1: 15+C=1\dfrac15 + C = 1, so C=45C = \dfrac45.

Substituting back confirms the result: y′=3t25−85t3y' = \dfrac{3t^2}{5} - \dfrac{8}{5t^3}, so t y′+2y=3t35−85t2+2t35+85t2=t3t\,y' + 2y = \dfrac{3t^3}{5} - \dfrac{8}{5t^2} + \dfrac{2t^3}{5} + \dfrac{8}{5t^2} = t^3, and y(1)=15+45=1y(1) = \dfrac15 + \dfrac45 = 1.

y(t)=t35+45t2y(t) = \dfrac{t^3}{5} + \dfrac{4}{5t^2}

Question 3Worked problemLaplace transform methods

Use the Laplace transform to solve

y′′+4y=8 u(t−π),y(0)=0,y′(0)=2,y'' + 4y = 8\,u(t - \pi), \qquad y(0) = 0, \quad y'(0) = 2,

where u(t−π)u(t - \pi) is the unit step function that equals 0 for t<πt < \pi and 1 for t≥πt \ge \pi. Express the solution as a piecewise function of tt.

Worked solution
Answer:

y(t)=sin⁡2ty(t) = \sin 2t for 0≤t<π0\le t<\pi, and y(t)=sin⁡2t+2−2cos⁡2ty(t) = \sin 2t + 2 - 2\cos 2t for t≥πt \ge \pi

Take the Laplace transform of both sides, using L{y′′}=s2Y(s)−s y(0)−y′(0)\mathcal{L}\{y''\} = s^2 Y(s) - s\,y(0) - y'(0) and the second shifting theorem L{u(t−a)}=e−ass\mathcal{L}\{u(t-a)\} = \dfrac{e^{-as}}{s}:

s2Y−0−2+4Y=8e−πss⟹Y(s)=2s2+4+e−πs 8s(s2+4).s^2 Y - 0 - 2 + 4Y = \frac{8e^{-\pi s}}{s} \quad\Longrightarrow\quad Y(s) = \frac{2}{s^2 + 4} + e^{-\pi s}\,\frac{8}{s(s^2+4)}.

The first term inverts directly: L−1{2s2+4}=sin⁡2t\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+4}\right\} = \sin 2t.

For the second, decompose 8s(s2+4)\dfrac{8}{s(s^2+4)} by partial fractions: 8s(s2+4)=As+Bs+Cs2+4\dfrac{8}{s(s^2+4)} = \dfrac{A}{s} + \dfrac{Bs + C}{s^2+4} gives A(s2+4)+(Bs+C)s=8A(s^2+4) + (Bs + C)s = 8; s=0s = 0 gives A=2A = 2, and matching the s2s^2 and ss coefficients gives B=−2B = -2, C=0C = 0. So

8s(s2+4)=2s−2ss2+4⟹L−1{8s(s2+4)}=2−2cos⁡2t=:g(t).\frac{8}{s(s^2+4)} = \frac{2}{s} - \frac{2s}{s^2+4} \quad\Longrightarrow\quad \mathcal{L}^{-1}\left\{\frac{8}{s(s^2+4)}\right\} = 2 - 2\cos 2t =: g(t).

The factor e−πse^{-\pi s} shifts and switches on this response at t=πt = \pi: L−1{e−πsG(s)}=u(t−π) g(t−π)\mathcal{L}^{-1}\{e^{-\pi s}G(s)\} = u(t-\pi)\,g(t-\pi). Since cos⁡(2(t−π))=cos⁡(2t−2π)=cos⁡2t\cos\big(2(t-\pi)\big) = \cos(2t - 2\pi) = \cos 2t,

y(t)=sin⁡2t+u(t−π)(2−2cos⁡2t).y(t) = \sin 2t + u(t-\pi)\left(2 - 2\cos 2t\right).

Piecewise:

y(t)={sin⁡2t,0≤t<π,sin⁡2t+2−2cos⁡2t,t≥π.y(t) = \begin{cases} \sin 2t, & 0 \le t < \pi, \\ \sin 2t + 2 - 2\cos 2t, & t \ge \pi. \end{cases}

Both pieces satisfy the differential equation (for t>πt > \pi: y′′+4y=(−4sin⁡2t+8cos⁡2t)+(4sin⁡2t+8−8cos⁡2t)=8y'' + 4y = (-4\sin 2t + 8\cos 2t) + (4\sin 2t + 8 - 8\cos 2t) = 8), the initial conditions hold, and yy and y′y' are continuous at t=πt = \pi because 2−2cos⁡2π=02-2\cos 2\pi = 0 and 4sin⁡2π=04\sin 2\pi = 0.

y(t)=sin⁡2ty(t) = \sin 2t for 0≤t<π0\le t<\pi, and y(t)=sin⁡2t+2−2cos⁡2ty(t) = \sin 2t + 2 - 2\cos 2t for t≥πt \ge \pi.

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