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.
Builds on Rates of change and building differential equations and Integrating standard functions.
Where it earns its keep: Radioactive decay and half-life on InkPhysics.
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.
COMMON MISCONCEPTION
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). Solve g(y) = 0 before dividing by it. Each root k of that gives a constant solution y = k, which satisfies the equation outright and which dividing by g(y) quietly deletes. Then divide by g(y) and every y goes to one side while every x goes to the other.
After that, both sides are ordinary integrals from earlier in this unit. One constant c, written the moment the integrals happen, covers the pair of them.
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 question said y > 0, and that is what lets the modulus go and keeps A positive here.
The condition gives A = 3, so y = 3ex²/2.
The constant cannot wait. Postpone it to the end and write y = ex²/2 + c and you have a function that fails the original equation, because the constant had to ride through the exponentiation and arrive as a multiplier.
GUIDED PRACTICE
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. Factorising fixes that: dy/dx = x(y + 1).
∫dy/(y + 1) = ∫x dx gives ln|y + 1| = x2/2 + c, so |y + 1| = ecex²/2.
Nothing here fixes the sign of y + 1, so the modulus comes off as a ±: y + 1 = Aex²/2 with A = ±ec, which is any non-zero constant. Taking A = ec alone would keep only y > −1 and lose the whole branch below it.
The condition gives A = 2, so y = 2ex²/2 − 1.
A = 0 is worth adding back. It returns y = −1, the constant solution that dividing by y + 1 threw away, and substituting it gives dy/dx = 0 and x(y + 1) = 0, so it does satisfy the equation. Allow every real A and y = Aex²/2 − 1 is the complete family: A > 0 above the line y = −1, A < 0 below it, A = 0 on it.
Spotting the common factor is the mark the whole question is built around. Everything after it is the previous example again, with one extra habit: ln|…| exponentiates to a constant of either sign, and the equilibrium solution is the zero case.
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 that forecast interpreted. What happens for large t? When does the model stop applying, and which of its assumptions gives way first? Those sentences carry their own marks, and students who can integrate perfectly still drop them.
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.
INDEPENDENT PRACTICE
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 stops things in finite time, so the exponential is trustworthy early and increasingly a fiction late. That is the kind of sentence the interpretation mark is looking for.
ASSESSMENT FOCUS
- 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, and questions expect that step written down. From ln|y| it is A = ±ec, any non-zero constant; only a stated restriction such as y > 0 makes A = ec positive.
- A = 0 puts the constant solution back into the family. Say so when the question asks for the general solution, since dividing by g(y) is what removed it.
- Particular solutions need the condition substituted into the general solution, never 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 holds only on −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.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the solving differential equations questions page.
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.
Open the full revision checklist to see every objective in the course in one place.