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Modelling with differential equations questions
Springs, dampers and linked populations are described by the same kind of equation. The roots of the auxiliary equation determine whether the behaviour is oscillation, damping or decay.
6 original questions · 25 marks · the modelling with differential equations notes · Differential equations
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In the oscillator model y'' + by' + cy = 0, state what the y' term and the y term each represent physically.
Worked answer
The y term is the restoring force pulling back towards equilibrium, its coefficient the stiffness; the y' term is resistance, draining energy in proportion to speed. B1 B1 for the two terms.Classify the damping in y'' + 8y' + 25y = 0, and state the frequency and decay rate of the motion.
Worked answer
Discriminant 64 − 100 = −36 < 0: light damping. The roots are −4 ± 3i, so the motion oscillates at angular frequency 3 inside a decaying envelope e−4t: frequency from the imaginary part, decay from the real part. M1 for the discriminant, A1 for light damping, B1 for the frequency and decay rate.A shock absorber obeys y'' + 10y' + 25y = 0. Classify the damping, give the general solution, and explain why this case is the design target for suspension.
Worked answer
Discriminant 100 − 100 = 0: critical damping, repeated root m = −5, solution y = (A + Bt)e−5t. Critical damping returns the system to rest as fast as possible with no overshoot; lighter damping bounces, heavier creeps. M1 for the discriminant, A1 for critical damping, A1 for the general solution, B1 for the design reason.Classify the damping in y'' + 5y' + 4y = 0 and describe the long-term motion.
Worked answer
Discriminant 25 − 16 = 9 > 0: heavy damping, real roots −1 and −4. The solution Ae−t + Be−4t creeps back to equilibrium without ever oscillating, the slower exponential setting the pace at late times. M1 for the discriminant, A1 for heavy damping with the roots, B1 for the long-term description.Solve the coupled system dx/dt = 2y, dy/dt = −2x with x(0) = 3 and y(0) = 0.
Worked answer
Differentiate the first equation: x'' = 2 dy/dt = −4x, so x'' + 4x = 0 and x = A cos 2t + B sin 2t. x(0) = 3 gives A = 3; y = x'/2 = −A sin 2t + B cos 2t at t = 0 gives B = 0. So x = 3 cos 2t and y = −3 sin 2t: the pair orbit a circle of radius 3 at angular speed 2. M1 for differentiating to eliminate y, A1 for x'' + 4x = 0, A1 for the general solution, M1 for applying both conditions, A1 for x and y.A mass on a spring is driven so that its displacement y at time t satisfies y'' + 2y' + 5y = 8cos t, with y = 0 and y' = 0 when t = 0. Find y in terms of t, and describe the motion after a long time.
Worked answer
Complementary function. The auxiliary equation m² + 2m + 5 = 0 has discriminant 4 − 20 = −16, so m = −1 ± 2i and the complementary function is e−t(A cos 2t + B sin 2t).
Particular integral. Try y = p cos t + q sin t. Then y′ = −p sin t + q cos t and y″ = −p cos t − q sin t. Substituting and collecting:
(−p + 2q + 5p)cos t + (−q − 2p + 5q)sin t = 8cos t, so 4p + 2q = 8 and 4q − 2p = 0.
From the second, p = 2q; then 8q + 2q = 8 gives q = 0.8 and p = 1.6.
General solution. y = e−t(A cos 2t + B sin 2t) + 1.6cos t + 0.8sin t.
Conditions. Apply them to the general solution, never to the complementary function alone; that is the error that ends this question. y(0) = 0 gives A + 1.6 = 0, so A = −1.6.
Differentiating, y′(0) = −A + 2B + 0.8, so 1.6 + 2B + 0.8 = 0 and B = −1.2.
Hence y = −e−t(1.6cos 2t + 1.2sin 2t) + 1.6cos t + 0.8sin t.
M1 for the auxiliary equation, A1 for the complementary function, M1 for a suitable particular integral, A1 for p and q, M1 for applying the conditions to the general solution, A1 for the value of A, A1 for the value of B, B1 for the steady state.
Long term. The e−t factor kills the first part, which is the transient. What remains is the steady state 1.6cos t + 0.8sin t: an oscillation at the driving frequency, of amplitude √(1.6² + 0.8²) = √3.2 ≈ 1.79. The system ends up oscillating at the frequency it is driven at, not at its own natural frequency of 2, and the initial conditions leave no trace.
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