Solving differential equations by separation of variables (WACE Year 12 Mathematics Specialist Unit 4)

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.

What a first-order differential equation is

A first-order differential equation relates a function to its first derivative, expressing the rate of change of a quantity in terms of the quantity itself and the independent variable. Solving it means finding the function whose derivative matches the equation, which reverses the differentiation and therefore brings in a constant of integration. Because of that constant, a single differential equation has a whole family of solutions, one for each starting value; the extra piece of information that selects one member of the family is the initial condition. Separation of variables is the standard technique for the important subclass of first-order equations that factor into a product of a function of $x$ and a function of $y$.

Recognising separability

An equation is separable when the right side factors into a part depending only on $x$ times a part depending only on $y$. Then all the $y$-dependence can be moved to one side with $dy$ and all the $x$-dependence to the other with $dx$.

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

Why treating the derivative as a fraction is valid

Writing $\dfrac{dy}{dx} = g(x)h(y)$ as $\dfrac{1}{h(y)}\,dy = g(x)\,dx$ looks like splitting a fraction, but the rigorous justification is the chain rule. Dividing by $h(y)$ gives $\dfrac{1}{h(y)}\dfrac{dy}{dx} = g(x)$, and integrating both sides with respect to $x$ gives $\displaystyle\int \dfrac{1}{h(y)}\dfrac{dy}{dx}\,dx = \displaystyle\int g(x)\,dx$. The left integral, by the substitution rule with $y$ as the new variable, is exactly $\displaystyle\int \dfrac{1}{h(y)}\,dy$. So the differential manipulation is a shorthand for a substitution, which is why it always produces the correct answer.

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 $x$ and $y$; rearrange for $y$ explicitly only if the algebra permits.

Which integration technique each side needs

The two integrals after separation can each demand any of the Unit 4 techniques, so a separable equation often combines skills. The right side, a function of $x$, may need a substitution or a trigonometric identity. The left side, $\displaystyle\int \dfrac{1}{h(y)}\,dy$, is frequently the harder one: when $h(y)$ is linear it gives a logarithm, but when $h(y)$ is a product such as $P(1 - P/M)$ in the logistic equation, it requires a partial-fraction decomposition before integrating. Recognising in advance which technique each side calls for keeps the working organised and avoids getting stuck halfway through a multi-step solution.

Applying an initial condition

Watching for lost solutions

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