Solve separable first-order differential equations of the form $\frac{dy}{dx} = f(x) g(y)$ by separating variables and integrating both sides

What this dot point is asking

NESA wants you to solve a first-order differential equation of the form $\frac{dy}{dx} = f(x) g(y)$ by separating the variables, integrating each side, and applying any given initial condition to find the constant of integration.

The conceptual leap is that $\frac{dy}{dx}$, though it is really a single limit, can be manipulated like a fraction here: you gather all the $y$-stuff with $dy$ on one side and all the $x$-stuff with $dx$ on the other, then integrate each side with respect to its own variable. This works precisely because the right-hand side factorises into a part in $x$ times a part in $y$ - which is what "separable" means.

The answer

The separation method

For an equation $\frac{dy}{dx} = f(x) g(y)$:

1. Rewrite as $\frac{dy}{g(y)} = f(x) \, dx$.
2. Integrate both sides: $\int \frac{dy}{g(y)} = \int f(x) \, dx + C$.
3. Solve for $y$ in terms of $x$ (or leave in implicit form if cleaner).
4. Apply any initial condition $y(x_0) = y_0$ to determine $C$.

If $g(y_0) = 0$ for some value $y_0$, then $y = y_0$ is a constant solution (an equilibrium of the equation). This must be checked separately, because the separation step divides by $g(y)$ and so silently throws it away.

Reading the equation geometrically, stage by stage

A first-order ODE assigns a gradient to every point of the plane, and a solution is a curve that follows those gradients. Seeing this picture makes the algebra meaningful: separating and integrating is just finding the curves that thread through the gradient field. The figures below use $\dfrac{dy}{dx} = -\dfrac{x}{y}$, whose solutions turn out to be circles.

Stage 1, the slope field. The equation gives a slope at every point: at $(x, y)$ the gradient is $-\dfrac{x}{y}$. Drawing a short segment of that slope at a grid of points produces the slope field, which already hints at a circular pattern.

Stage 2, separate and integrate. Gather the variables: $y\,dy = -x\,dx$. Integrating both sides gives $\tfrac12 y^2 = -\tfrac12 x^2 + C$, which tidies to $x^2 + y^2 = C$. The algebra has produced the equation of the curves the slope field was tracing.

Stage 3, the general solution. $x^2 + y^2 = C$ is a family of concentric circles, one for each value of the constant $C > 0$. Every member of the family is a solution; the slope-field segments are tangent to these circles, which is the visual meaning of "the curve follows the gradients".

Stage 4, apply the initial condition. A condition such as $y = 2$ when $x = 0$ fixes the constant: $0 + 4 = C$, so $C = 4$. That selects the single circle $x^2 + y^2 = 4$ from the family - the particular solution through the given point.

When separation is possible

A first-order ODE separates if and only if you can write $\frac{dy}{dx} = f(x) g(y)$ for some functions $f$ and $g$. Examples that separate:

• $\frac{dy}{dx} = x y$.
• $\frac{dy}{dx} = e^x \sin y$.
• $\frac{dy}{dx} = \frac{y + 1}{x - 1}$.

Examples that do not separate:

• $\frac{dy}{dx} = x + y$ (a sum, not a product).
• $\frac{dy}{dx} = \frac{y}{x + y}$ (cannot be factored cleanly).

Extension 1 problems are always separable; non-separable equations belong to Extension 2.

The constant of integration

Always include a constant after integration. It is cleaner to combine the constants from both sides into one $C$ on the right.

If the question gives an initial condition, substitute to find $C$. If not, leave the answer in terms of $C$. A favourite simplification is to relabel a messy constant: after exponentiating $\ln|y| = x^3 + C$, the factor $e^C$ becomes a single new constant $A$, and you should say so explicitly.

Newton's law of cooling

A body cools toward an ambient temperature $T_a$ at a rate proportional to the temperature difference:

where $k > 0$.

Separate: $\frac{dT}{T - T_a} = -k \, dt$.

Integrate: $\ln |T - T_a| = -k t + C$.

So $T - T_a = A e^{-k t}$ for some constant $A$, giving

The temperature approaches $T_a$ as $t \to \infty$. Note that $T = T_a$ is the equilibrium (constant) solution, the case $A = 0$: a body already at room temperature stays there.

Implicit versus explicit solutions

Sometimes the integrated equation cannot be solved cleanly for $y$. In that case, leave it implicit. For example,

This is the family of circles $x^2 + y^2 = C$, the implicit solution - and the one drawn in the stage figures above.

How exam questions ask about separable ODEs

• "Solve the differential equation ..." (with an initial condition). Separate, integrate, apply the condition to find the constant, and give $y$ explicitly (choosing the correct sign of any root from the initial condition).
• "Find the general solution." Same, but stop at the family with $+C$; no initial condition is given.
• "Show that $y = \ldots$ satisfies / is a solution." You can verify by differentiating, but a "solve" instruction wants the separation method shown.
• "A body cools / a population grows ..." A modelling problem: set up the proportionality as a differential equation, separate and solve, then use the data points to find the constants $k$ and $A$.
• "Find the constant solution / equilibrium." Set $\frac{dy}{dx} = 0$, i.e. solve $g(y) = 0$; this is the solution the separation step hides.
• "Leave your answer in implicit form." Do not force an explicit $y$; the integrated equation is the answer.