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Solving differential equations questions

The rates lesson taught how to write a differential equation; this one teaches how to solve it. Separable equations split so that y-things face x-things across the equals sign, both sides integrate, and a known point turns the family of solutions into the single curve a model actually needs.

6 original questions · 28 marks · the solving differential equations notes · Integration

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

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

  Worked 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. The separation is bookkeeping, and the calculus starts once it is done.

• 4 marks · warm-up

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

  Worked 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. One constant, absorbed once and pinned once. Writing y = e^x³ + c instead of absorbing the constant into a multiplier is the error that costs the accuracy marks.

• 5 marks

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

  Worked 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. A tame-looking equation has produced a vertical asymptote, and the validity statement carries its own mark.

• 5 marks

  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.

  Worked 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. Unlike an exponential decay, this tank genuinely empties in finite time. Note that the model only describes the drain up to t = 8/k, since beyond that the formula would have the depth rising again.

• 5 marks

  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.

  Worked 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.

• 7 marks · stretch

  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).

  Worked 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. Three places lose marks. 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. And the final division by e^0.5t is needed to match the printed answer, so stopping one line early forfeits the last mark on a show-that.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise solving differential equations one question at a time

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