A-level Mathematics

7.10 Constructing differential equations — Practise with help

Work each question closed book first. Read the sentence twice: once to find the rate, and once to find the sign.

Warm-up: remember

  1. Write " yy is proportional to x2x^2 " as an equation.
  2. Make kk the subject of 5=2k5 = 2k.
  3. Differentiate y=Ae3xy = Ae^{3x} with respect to xx.
Answers
  1. y=kx2y = kx^2, where kk is a constant.
  2. k=52=2.5k = \dfrac{5}{2} = 2.5.
  3. dydx=3Ae3x\dfrac{dy}{dx} = 3Ae^{3x}.

If question 1 was hard, revise proportionality in Chapter 2.

Practise with help

1. A population PP of insects grows at a rate proportional to PP. Write a differential equation relating dPdt\dfrac{dP}{dt} to PP.

Hint

"Proportional to PP" introduces a constant kk. The population grows, so the sign is positive.

Next step

Write the rate on the left and the proportional expression on the right: dPdt\dfrac{dP}{dt} equals a constant times PP.

Full working dPdt=kP,k>0.\frac{dP}{dt} = kP, \qquad k > 0.

The larger the population, the faster it grows, because both sides grow with PP.

2. A radioactive sample of mass mm grams decays at a rate proportional to its mass. Write a differential equation for mm after tt years.

Hint

Decay means the mass is falling, so the rate is negative. "Proportional to its mass" introduces a constant kk.

Next step

Write dmdt\dfrac{dm}{dt} equal to minus a constant times mm.

Full working dmdt=−km,k>0.\frac{dm}{dt} = -km, \qquad k > 0.

The minus sign says the mass decreases. The constant kk is the decay constant, and a larger kk means faster decay.

3. A metal block cools in air at 15∘15^\circC. Its temperature θ\theta falls at a rate proportional to the difference between θ\theta and the air temperature. Write a differential equation for θ\theta after tt minutes.

Hint

The temperature difference is θ−15\theta - 15. The block cools, so the rate is negative while θ>15\theta > 15.

Next step

Write dθdt=−k(θ−15)\dfrac{d\theta}{dt} = -k(\theta - 15).

Full working dθdt=−k(θ−15),k>0.\frac{d\theta}{dt} = -k(\theta - 15), \qquad k > 0.

When θ>15\theta > 15 the right side is negative, so the temperature falls. As θ\theta approaches 1515 the right side approaches zero, so the cooling slows down.

4. A car accelerates at a constant rate aa. Its velocity is vv after tt seconds. Write a differential equation for vv.

Hint

Acceleration is the rate of change of velocity.

Next step

Write the rate of change of vv equal to the constant aa.

Full working dvdt=a.\frac{dv}{dt} = a.

A constant acceleration means the velocity changes at a steady rate, so the graph of vv against tt is a straight line of gradient aa.

5. The curve y=Asin⁡2xy = A\sin 2x contains a constant AA. Form a differential equation that does not contain AA.

Hint

Differentiate twice. The second derivative brings back −sin⁡2x-\sin 2x, which is proportional to yy.

Next step

dydx=2Acos⁡2x\dfrac{dy}{dx} = 2A\cos 2x, so d2ydx2=−4Asin⁡2x\dfrac{d^2y}{dx^2} = -4A\sin 2x. Now replace Asin⁡2xA\sin 2x with yy.

Full working dydx=2Acos⁡2x,d2ydx2=−4Asin⁡2x=−4y.\frac{dy}{dx} = 2A\cos 2x, \qquad \frac{d^2y}{dx^2} = -4A\sin 2x = -4y.

The differential equation is d2ydx2=−4y\dfrac{d^2y}{dx^2} = -4y. Every curve in the family satisfies it, whatever the value of AA.

6. The curve y=x2+Cy = x^2 + C contains a constant CC. Form a differential equation that does not contain CC.

Hint

Differentiate once and see whether the constant survives.

Next step

dydx=2x\dfrac{dy}{dx} = 2x. The constant CC has already gone.

Full working dydx=2x.\frac{dy}{dx} = 2x.

Every curve y=x2+Cy = x^2 + C has the same gradient at a given xx, because adding a constant moves the curve up or down without changing its slope.

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