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WASpecialist MathematicsSyllabus dot point

When a differential equation factors into a function of x times a function of y, how do we solve it by separating the variables?

Solve separable first-order differential equations and apply an initial condition to find the particular solution

WACE Specialist Unit 4 separation of variables: recognising a separable equation, moving all y terms to one side and x terms to the other, integrating both sides, and using an initial condition to fix the constant, with a worked example.

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  1. What this dot point is asking
  2. Recognising separability
  3. Integrating both sides
  4. Applying an initial condition
  5. Watching for lost solutions

What this dot point is asking

SCSA wants you to recognise a separable equation, carry out the separation and integration carefully, and apply a given condition to find the constant.

Recognising separability

An equation is separable when the right side factors into a part depending only on xx times a part depending only on yy. Then all the yy-dependence can be moved to one side with dydy and all the xx-dependence to the other with dxdx.

Treating dydx\tfrac{dy}{dx} as a ratio of differentials is a valid shortcut here; the underlying justification is the chain rule.

Integrating both sides

Integrate each side with respect to its own variable. A single arbitrary constant suffices (combine the two constants of integration into one). The result is often an implicit relation between xx and yy; rearrange for yy explicitly only if the algebra permits.

Applying an initial condition

Watching for lost solutions

Dividing by h(y)h(y) assumes h(y)0h(y) \neq 0. Any value of yy making h(y)=0h(y) = 0 is a constant equilibrium solution that the separation step can hide; note it separately.