Maths › Integration › Solving differential equations

Solving differential equations

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.

Year 12-13EDEXCEL 9MA0 8.7, 8.8

Builds on Rates of change and building differential equations and Integrating standard functions.

IN THIS TOPIC

  • Solve separable equations, factorising first where separation needs it.
  • Turn a general solution into a particular one with a known condition.
  • Interpret solutions in context and state the limits of what the model can say.

WHAT YOU PROBABLY THINK

Integrate both sides, rearrange, and add the + c at the end.

Separate, integrate, solve

A separable equation has the shape dy/dx = f(x)g(y), and dividing by g(y) sends every y to one side while dx-scaling sends every x to the other,

∫ 1g(y) dy = ∫ f(x) dx

after which both sides are ordinary integrals from earlier in this unit. One constant c, written the moment the integrals happen, covers both sides.

WORKED EXAMPLE

The full routine

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

Separate and integrate: ∫(1/y) dy = ∫x dx, so ln y = x2/2 + c.

Exponentiate: y = Aex²/2, where A = ec absorbs the constant.

The condition gives A = 3, so y = 3ex²/2.

Here the top-of-page lie collapses. Waiting until the end and writing y = ex²/2 + c gives a function that fails the original equation, because the constant had to ride through the exponentiation, arriving as a multiplier, not an afterthought.

The family of solution curves y equals A times e to the x squared over 2 for four values of A, every member satisfying d y by d x equals x y, with the point 0 comma 3 marked on the one member that passes through it(0, 3) picks A = 3A = ½, 1, 2, 3one equation, a family of curves: y = Ae^(x²/2)
FIG. 1The general solution y = Ae^(x²/2) is a family, one curve per A. The condition y(0) = 3 picks out the amber member.

YOUR TURN

Factorise before separating

Solve dy/dx = xy + x, given that y = 1 when x = 0, before opening the working.

Show the working

The right side does not separate as written, but factorising does it: dy/dx = x(y + 1).

∫dy/(y + 1) = ∫x dx gives ln|y + 1| = x2/2 + c, so y + 1 = Aex²/2.

The condition gives A = 2, so y = 2ex²/2 − 1.

Spotting the common factor is the mark the question is built around; everything after it is the previous example again.

Models with meaning

In context, a differential equation is a claim about how the world changes, and its solution is a forecast. The exam wants the forecast interpreted: what happens for large t, when the model stops applying, and which of its assumptions gives way first.

WORKED EXAMPLE

The dissolving mint

A spherical mint's radius decreases at a rate inversely proportional to the square of the radius: dr/dt = −k/r2, with r = 1 cm at t = 0. Solve for r.

Separate: ∫r2 dr = ∫−k dt, so r3/3 = −kt + c.

The condition gives c = 1/3, so r3 = 1 − 3kt.

With k = 0.01, r reaches zero at t = 100/3 ≈ 33 minutes, and there the model ends: the equation would happily push r negative, but the mint has already gone. A solution's validity window is part of the answer.

The radius of a dissolving mint against time, falling from 1 along r cubed equals 1 minus 0.03 t and meeting zero near t equals 33, where the model stops meaning anythingr³ = 1 − 0.03tmodel stops: r = 00100/3dr/dt = −k/r²: solved, then interpreted
FIG. 2The mint's radius against time: r³ falls linearly, the sphere vanishes at t = 100/3, and beyond the dashed line the algebra describes nothing.

TRY IT UNSEEN

Slowing down

A particle moving in a straight line decelerates so that dv/dt = −0.2v, with v = 20 m s⁻¹ at t = 0. Find v(t), and comment on what the model says for large t.

Show the working

Separating, ∫dv/v = ∫−0.2 dt gives ln v = −0.2t + c, so v = Ae−0.2t with A = 20.

v = 20e−0.2t, halving roughly every 3.5 seconds.

For large t the model has v approach zero without ever reaching it. Real friction brings things to rest in finite time, so the exponential is trustworthy early and increasingly a fiction late, which is the kind of sentence the interpretation mark wants.

THE EXAM BIT

  • Separate fully before integrating, factorising out any common factor first, and show the separated line as working.
  • One + c, written when the integrals happen; a constant bolted on after rearranging fails the original equation.
  • Through an exponentiation the constant becomes a multiplier A = e^c, and questions expect that step shown.
  • Particular solutions need the condition substituted into the general solution, not into the differential equation.
  • In context, state the validity window and the large-t behaviour in a sentence; the interpretation carries its own mark.

CHECK YOURSELF

Solve dy/dx = y2 sin x, given y = 1 when x = 0, and state where the solution stops being valid.

Show a hint

∫y⁻² dy = ∫sin x dx, then the condition; watch the denominator.

Show the answer

Separating, −1/y = −cos x + c, and the condition y(0) = 1 gives c = 0.

So 1/y = cos x, that is y = 1/cos x.

The solution blows up where cos x = 0, so it is valid only for −π/2 < x < π/2 around the starting point; a solved model can still carry an expiry date.

Separate y-things from x-things, integrate both sides, and let one constant cover it.

The constant boards at integration time, and in context the solution comes with a validity window.

CHECK YOUR PROGRESS

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  • Solve separable equations, factorising first where separation needs it.
  • Turn a general solution into a particular one with a known condition.
  • Interpret solutions in context and state the limits of what the model can say.

No animated video for this topic yet; these notes stand alone.