What are two separate constants?

Use a single combined constant; do not carry one on each side.

SCSA WACE 2022-style practice: Calculator-assumed. A tank's water height h (cm) satisfies dhdt = -2sqrt(h), with h = 100 when t = 0. (a) Solve for h as a function of t. (b) Find the time at which the tank is empty.

A separable model with a square root. (a) Separate: h^-1/2\,dh = -2\,dt. Integrate: 2h^1/2 = -2t + C. At t = 0, h = 100: 2sqrt(100) = C, so C = 20. Thus 2sqrt(h) = 20 - 2t, giving sqrt(h) = 10 - t and h = (10 - t)^2 for 0 t 10. (b) The tank is empty when h = 0: (10 - t)^2 = 0, so t = 10 seconds. Markers reward the separation, integrating h^-1/2 to 2h^1/2, applying the initial condition, and the empty time t = 10.

§-Syllabus dot point

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.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
What a first-order differential equation is
Recognising separability
Why treating the derivative as a fraction is valid
Integrating both sides
Which integration technique each side needs
Applying an initial condition
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.

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.

Worked example

Solve $\dfrac{dy}{dx} = xy$ with $y = 2$ when $x = 0$ .

Step 1: Separate the variables

We rearrange so that all terms involving $y$ are on the left and all terms involving $x$ are on the right, which allows each side to be integrated independently. Dividing by $y$ is valid when $y \ne 0$ ; note that $y = 0$ is a separate equilibrium solution that the division step would lose.

\frac{1}{y}\,dy = x\,dx.

Step 2: Integrate both sides

Each side now depends on only one variable, so we integrate directly and absorb the two constants of integration into a single arbitrary constant $C$ .

\ln|y| = \frac{x^2}{2} + C.

Step 3: Solve for $y$ in explicit form

Exponentiating both sides removes the logarithm. Writing $A = \pm e^{C}$ captures all non-zero solutions in a single expression.

y = A\,e^{x^2/2}.

Step 4: Apply the initial condition

Substituting the known point $y = 2$ at $x = 0$ determines the constant $A$ for this particular solution.

2 = A\,e^{0} = A, \quad \text{so } A = 2.

Final answer: $y = 2\,e^{x^2/2}$ .

Exam-style practice questions

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

WACE 20227 marksCalculator-assumed. A tank's water height

h

(cm) satisfies

\dfrac{dh}{dt} = -2\sqrt{h}

, with

h = 100

when

t = 0

. (a) Solve for

h

as a function of

t

. (b) Find the time at which the tank is empty.

Show worked answer →

A separable model with a square root.

(a) Separate: $h^{-1/2}\,dh = -2\,dt$ . Integrate: $2h^{1/2} = -2t + C$ . At $t = 0$ , $h = 100$ : $2\sqrt{100} = C$ , so $C = 20$ . Thus $2\sqrt{h} = 20 - 2t$ , giving $\sqrt{h} = 10 - t$ and $h = (10 - t)^2$ for $0 \le t \le 10$ .

(b) The tank is empty when $h = 0$ : $(10 - t)^2 = 0$ , so $t = 10$ seconds.

Markers reward the separation, integrating $h^{-1/2}$ to $2h^{1/2}$ , applying the initial condition, and the empty time $t = 10$ .

WACE 20215 marksCalculator-free. Solve

\dfrac{dy}{dx} = \dfrac{\cos x}{y}

given that

y = 2

when

x = 0

Show worked answer →

A trig-on-the-right separable equation.

Separate: $y\,dy = \cos x\,dx$ . Integrate both sides: $\dfrac{y^2}{2} = \sin x + C$ , so $y^2 = 2\sin x + 2C$ . Write $2C = K$ : $y^2 = 2\sin x + K$ .

Apply $y = 2$ at $x = 0$ : $4 = 2(0) + K$ , so $K = 4$ . Thus $y^2 = 2\sin x + 4$ , and since $y = 2 > 0$ , $y = \sqrt{2\sin x + 4}$ .

Markers reward the separation, integrating both sides, applying the condition, and the positive square root for $y$ .