Questions on solving differential equations

Question 1

Separate the variables in dy/dx = x²/y, writing the result as two integrals.

Accepted Answer

Multiply through by y and integrate both sides to reach ∫y dy = ∫x² dx. Everything in y moves left with dy, everything in x stays right with dx. M1 for separating the variables, A1 for the two integrals.

Question 2

Solve dy/dx = 3x²y for y > 0, given that y = 2 when x = 0.

Accepted Answer

Separate to get ∫(1/y) dy = ∫3x² dx, so ln y = x³ + c. Exponentiating gives y = Ae^x³, where A = e^c. The condition gives A = 2, so y = 2e^x³. M1 for separating, A1 for ln y = x³ + c, M1 for exponentiating, A1 for y = 2e^x³. The constant enters as a multiplier, y = Ae^x³, because it is added before the exponential is taken.

Question 3

Separate the variables in dy/dx = e^x − y, writing the result as two integrals.

Accepted Answer

The right-hand side is one exponential rather than a product, but a law of indices splits it: e^x − y = e^xe^−y = e^x/e^y, a function of x over a function of y. M1 for the split. Now multiply both sides by e^y and by dx, which puts everything in y on the left with dy and everything in x on the right with dx. M1 for separating. The result is ∫e^y dy = ∫e^x dx. A1 for the two integrals. The step that separates is a multiplication by e^y, which is never zero, so no solution is lost along the way, unlike a division by a bracket that can vanish.

Question 4

Show that y = 4e^−2x + 3 satisfies dy/dx = 6 − 2y and the condition y = 7 when x = 0.

Accepted Answer

Differentiate: dy/dx = 4 × (−2)e^−2x = −8e^−2x. M1 for the derivative. Now work out the right-hand side from the proposed y: 6 − 2(4e^−2x + 3) = 6 − 8e^−2x − 6 = −8e^−2x. The two sides agree for every x, so the equation is satisfied. A1 for both sides reaching −8e^−2x. At x = 0, e⁰ = 1, so y = 4 + 3 = 7, and the condition holds. B1 for the check at x = 0.

Question 5

Solve dy/dx = y², given that y = 1 when x = 0, and state the values of x for which your solution is valid.

Accepted Answer

Separate to get ∫y⁻² dy = ∫dx, so −1/y = x + c. The condition gives c = −1, and rearranging gives y = 1/(1 − x). The solution grows without bound as x → 1, so it is valid only for x < 1. M1 for separating, A1 for −1/y = x + c, M1 for using y = 1 at x = 0, A1 for y = 1/(1 − x), B1 for x < 1.

Question 6

A tank is draining, and its depth h at time t satisfies dh/dt = −k√h, where k is a positive constant. Given that h = 16 when t = 0, solve the equation and find, in terms of k, the time at which the tank empties.

Accepted Answer

Separate to get ∫h^−1/2 dh = −k∫dt, so 2√h = −kt + c. The condition gives c = 8, hence √h = (8 − kt)/2 and h = (8 − kt)²/4. The depth reaches zero when kt = 8, that is at t = 8/k. M1 for separating, A1 for 2√h = −kt + c, M1 for c = 8, A1 for h = (8 − kt)²/4, A1 for t = 8/k. Note that the model only describes the drain up to t = 8/k, since beyond that the formula would have the depth rising again.

Question 7

Water leaks from a tank at a rate proportional to the volume present, so that dV/dt = −kV. Initially the tank holds 500 litres, and after 10 minutes it holds 250 litres. Find V as a function of t, and find the value of k to 3 significant figures.

Accepted Answer

Separating and integrating gives ln V = −kt + c, so V = 500e^−kt once the initial condition is applied. Then 250 = 500e^−10k gives e^−10k = ½, so 10k = ln 2 and k = (ln 2)/10 = 0.0693 per minute. M1 for separating, A1 for ln V = −kt + c, A1 for V = 500e^−kt, M1 for 250 = 500e^−10k, A1 for k = 0.0693. The ten-minute halving is this model's half-life, and every further ten minutes halves the volume again. Taking logarithms of ½ and mislaying the minus sign is the usual route to a negative k, which no leaking tank can have.

Question 8

Find the general solution of d²y/dx² + 5(dy/dx) + 6y = 0, and hence find the solution for which y = 5 and dy/dx = −12 when x = 0.

Accepted Answer

The auxiliary equation is m² + 5m + 6 = 0, which factorises as (m + 2)(m + 3) = 0, so m = −2 or m = −3. Two distinct real roots give the general solution y = Ae^−2x + Be^−3x.
Differentiating, dy/dx = −2Ae^−2x − 3Be^−3x. The conditions at x = 0 give A + B = 5 and −2A − 3B = −12.
From the first, A = 5 − B. Substituting, −2(5 − B) − 3B = −12, so −10 − B = −12 and B = 2, hence A = 3.
The particular solution is y = 3e^−2x + 2e^−3x.
M1 for forming the auxiliary equation, A1 for the roots −2 and −3, A1 for the general solution, M1 for differentiating and forming two simultaneous equations, A1 for the particular solution. Both roots are negative, so every solution of this equation decays to zero. The second condition is on dy/dx, so the general solution has to be differentiated before it can be used.

Question 9

Solve d²y/dx² − 6(dy/dx) + 9y = 0, given that y = 2 and dy/dx = 5 when x = 0. Explain why Ae^3x + Be^3x is not an acceptable general solution.

Accepted Answer

The auxiliary equation m² − 6m + 9 = 0 is (m − 3)² = 0, so m = 3 is a repeated root. The general solution is y = (A + Bx)e^3x.
Ae^3x + Be^3x collapses to (A + B)e^3x, which is a single arbitrary constant. A second order equation needs two independent arbitrary constants, and the second independent solution is xe^3x.
Differentiating by the product rule, dy/dx = Be^3x + 3(A + Bx)e^3x. At x = 0, y = 2 gives A = 2, and dy/dx = 5 gives B + 3A = 5, so B = −1.
Hence y = (2 − x)e^3x.
M1 for the auxiliary equation and its repeated root, A1 for the form (A + Bx)e^3x, B1 for the independence argument, M1 for the product rule and the two conditions, A1 for y = (2 − x)e^3x. The factor (2 − x) changes sign at x = 2, so this solution crosses the axis once before the exponential takes over.

Question 10

Find the general solution of dy/dx = 2x(y − 3), and state the solution that the separation removes.

Accepted Answer

Separate: ∫1/(y − 3) dy = ∫2x dx. M1 for separating. Integrating both sides gives ln|y − 3| = x² + c. A1 for that form, with one constant on one side. Exponentiating, |y − 3| = e^ce^x², and absorbing the sign and the e^c into a single constant A gives y − 3 = Ae^x². M1 for exponentiating and absorbing the constant. So the general solution is y = 3 + Ae^x². A1 for the general solution. Dividing by y − 3 at the first step assumed y ≠ 3, and y = 3 is itself a solution: both sides are then zero. B1 for y = 3, recovered by taking A = 0.

Question 11

Solve dy/dx = y cos x, given that y = 2 when x = 0, and describe the behaviour of y for large x.

Accepted Answer

Separate: ∫1/y dy = ∫cos x dx. M1 for separating. Integrating, ln|y| = sin x + c. A1 for that form. Exponentiating gives y = Ae^sin x, with A absorbing e^c and the sign. M1 for exponentiating. At x = 0, sin 0 = 0, so 2 = A and y = 2e^sin x. A1 for the particular solution. For large x the solution does not settle to anything. Since sin x runs between −1 and 1 with period 2π, y oscillates between 2/e ≈ 0.736 and 2e ≈ 5.44, repeating every 2π and approaching no limit, reaching the larger value at x = π/2 and the smaller at x = 3π/2. B1 for the bounded periodic behaviour with the two extremes.

Question 12

A population of P thousand insects at time t years is modelled by dP/dt = 0.1P(5 − P), 0 < P < 5. Given that P = 1 when t = 0, show that P = 5/(1 + 4e^−0.5t).

Accepted Answer

Separate to get ∫1/[P(5 − P)] dP = ∫0.1 dt. The left-hand integrand needs partial fractions: 1/[P(5 − P)] = (1/5)[1/P + 1/(5 − P)]. Integrating gives (1/5)[ln P − ln(5 − P)] = 0.1t + c, so ln[P/(5 − P)] = 0.5t + 5c. The condition P = 1 at t = 0 gives ln(1/4) = 5c, so ln[P/(5 − P)] = 0.5t − ln 4 and P/(5 − P) = ¼e^0.5t. Cross-multiplying, 4P = (5 − P)e^0.5t, so P(4 + e^0.5t) = 5e^0.5t and P = 5e^0.5t/(4 + e^0.5t). Dividing top and bottom by e^0.5t gives the required form. M1 for separating, M1 for partial fractions, A1 for the correct pair, A1 for the integrated form, M1 for using P = 1 at t = 0, A1 for P/(5 − P) = ¼e^0.5t, A1 for the printed form. Two steps need care. The 1/(5 − P) term integrates to −ln(5 − P), not +ln(5 − P), because of the inner coefficient of −1. The factor of 1/5 must be cleared before exponentiating.

Question 13

A mass on a spring is held 3 cm from its equilibrium position and released from rest when t = 0 seconds. Its displacement x cm at time t seconds is modelled by d²x/dt² + 4(dx/dt) + 13x = 0. Find x in terms of t, state the period of the oscillation, and describe the motion as t increases. State one assumption the model makes.

Accepted Answer

The auxiliary equation is m² + 4m + 13 = 0, with discriminant 16 − 52 = −36. So m = (−4 ± 6i)/2 = −2 ± 3i, and the general solution is x = e^−2t(A cos 3t + B sin 3t).
Released from rest means x = 3 and dx/dt = 0 at t = 0. The first gives A = 3. Differentiating, dx/dt = −2e^−2t(A cos 3t + B sin 3t) + e^−2t(−3A sin 3t + 3B cos 3t), so at t = 0 this is −2A + 3B = 0, giving B = 2.
Hence x = e^−2t(3 cos 3t + 2 sin 3t).
The oscillating part has angular frequency 3, so the period is 2π/3 ≈ 2.09 seconds.
The motion is a decaying oscillation: the displacement continues to alternate in sign with that fixed period, but the envelope √13 e^−2t shrinks, so the mass settles towards equilibrium. This is light damping, the case the negative discriminant identifies.
An assumption: the resistive force is taken to be proportional to the speed, with no dry friction and no external driving force, and the spring is assumed to obey Hooke's law throughout, which for a real spring holds only over a limited range of extension.
M1 for the auxiliary equation, A1 for the roots −2 ± 3i, A1 for the general solution, M1 for differentiating and applying both conditions, A1 for A = 3 and B = 2, B1 for the period 2π/3, B1 for the decaying-oscillation description with a stated assumption. The real part of the root controls the decay and the imaginary part controls the frequency, so the period does not depend on how far the mass was pulled.

Question 14

A population of P thousand fish in a lake is modelled by dP/dt = 0.2P − 5, where t is the time in years, and P = 20 when t = 0. Find P in terms of t, find the time at which the model says the population reaches zero, and state the interval of t on which the model can be used.

Accepted Answer

Factorise the right-hand side before separating: 0.2P − 5 = 0.2(P − 25). Separating, ∫1/(P − 25) dP = ∫0.2 dt. M1 for factorising and separating. The population starts at 20 and the equation makes dP/dt negative there, so P stays below 25 throughout and the modulus resolves as |P − 25| = 25 − P. Integrating gives ln(25 − P) = 0.2t + k, and differentiating ln(25 − P) returns 1/(P − 25), which is the check on that sign. A1 for the integrated form with the modulus resolved. At t = 0, 25 − 20 = 5, so k = ln 5. M1 for using P = 20 at t = 0. Hence 25 − P = 5e^0.2t and P = 25 − 5e^0.2t. A1 for the solution. Setting P = 0 gives 5e^0.2t = 25, so e^0.2t = 5 and 0.2t = ln 5. M1 for solving P = 0. The population reaches zero at t = 5 ln 5 ≈ 8.05 years, and the model can only be used for 0 ≤ t ≤ 5 ln 5, since beyond that it returns a negative population, which nothing it describes can have. A1 for the time with the interval. The subtraction of 5 is a constant harvest, and it is what turns a growth model into one that empties the lake: with P below 25 thousand the harvest exceeds the growth at every moment.

Question 15

Find the general solution of dy/dx = −x/y, describe the family of solution curves, and find the particular solution through (3, 4), stating the values of x for which it is valid.

Accepted Answer

Separate: ∫y dy = ∫−x dx. M1 for separating. Integrating both sides gives y²/2 = −x²/2 + c. A1 for that form. Multiplying by 2 and writing k for 2c, the general solution is x² + y² = k, left in implicit form. A1 for the implicit general solution. The solution curves are therefore circles centred at the origin, one for each k > 0, of radius √k. B1 for the family described. Through (3, 4), k = 9 + 16 = 25, so x² + y² = 25. M1 for the particular solution. A solution has to be a function of x, and the given point has y = 4, so it is the upper semicircle y = √(25 − x²), valid for −5 < x < 5. The endpoints are excluded because y = 0 there and −x/y is not defined, so the gradient the equation asks for does not exist at either end. B1 for the interval with that reason.