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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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 times a part depending only on . Then all the -dependence can be moved to one side with and all the -dependence to the other with .
Treating 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 and ; rearrange for explicitly only if the algebra permits.
Applying an initial condition
Watching for lost solutions
Dividing by assumes . Any value of making is a constant equilibrium solution that the separation step can hide; note it separately.