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SASpecialist MathematicsSyllabus dot point

How do we solve a differential equation by separating the two variables onto opposite sides?

Solve first-order separable differential equations and find particular solutions using initial conditions.

Solving first-order separable differential equations by separating variables and integrating both sides, applying initial conditions to find the particular solution, and handling the constant of integration.

Generated by Claude Opus 4.78 min answer

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  1. What this dot point is asking
  2. The separation method
  3. A worked general-then-particular solution
  4. Exponential growth and decay

What this dot point is asking

You need to solve first-order separable differential equations and use initial conditions to find particular solutions.

The separation method

If dydx=f(x) g(y)\dfrac{dy}{dx} = f(x)\,g(y), divide by g(y)g(y) and multiply by dxdx to gather like variables:

1g(y) dy=f(x) dx.\frac{1}{g(y)}\,dy = f(x)\,dx.

Integrate both sides:
∫1g(y) dy=∫f(x) dx.\int \frac{1}{g(y)}\,dy = \int f(x)\,dx.

A single constant of integration CC on one side captures the whole family of solutions.

A worked general-then-particular solution

Exponential growth and decay

The equation dydt=ky\dfrac{dy}{dt} = ky is separable and central to modelling:

∫1y dy=∫k dt  ⟹  ln⁡∣y∣=kt+C  ⟹  y=Aekt,\int \frac{1}{y}\,dy = \int k\,dt \implies \ln|y| = kt + C \implies y = A e^{kt},

where A=eCA = e^{C} is set by the initial value. Positive kk gives growth; negative kk gives decay. This is the backbone of population, radioactive-decay and cooling models.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2017 SACE Stage 23 marksThe rate of change of the area of a shadow is dA/dt = -2t + 150 cos(t) sin^2(t), where A is in square metres and t is the time in hours. At sunrise t = 0 and A = 375 m^2. Solve the differential equation, and hence find an expression for the area of the shadow at time t.
Show worked answer →

The variables are already separated (the right-hand side depends only on t), so integrate both sides with respect to t.

integral of -2t dt = -t^2.
For integral of 150 cos(t) sin^2(t) dt, use the substitution u = sin(t), du = cos(t) dt, giving 150 integral of u^2 du = 150 u^3 / 3 = 50 sin^3(t). [1 mark]

So A = -t^2 + 50 sin^3(t) + C, where C is the constant of integration. [1 mark]

Apply the initial condition A = 375 when t = 0: 375 = -(0)^2 + 50 sin^3(0) + C = C, so C = 375.

Therefore A = -t^2 + 50 sin^3(t) + 375. [1 mark]

2025 SACE Stage 22 marksThe rate of change of unreacted mass A is given by dA/dt = (1/4) dX/dt, where t is in minutes. Use integration to show that the solution is A = (1/4) X + 2, given that A = 2 when X = 0.
Show worked answer →

Treat dA/dt = (1/4) dX/dt as a relationship between the rates. Integrating both sides with respect to t (or equivalently dA = (1/4) dX) gives A = (1/4) X + C, where C is a constant. [1 mark]

Apply the condition A = 2 when X = 0: 2 = (1/4)(0) + C, so C = 2.

Therefore A = (1/4) X + 2, as required. [1 mark]