Separable Differential Equations

What does "separable" mean?

A first-order ODE is separable when you can rearrange it so that every $y$-term (including $dy$) sits on one side and every $x$-term (including $dx$) sits on the other: $\frac{dy}{dx} = f(x) \, g(y) \quad \Longleftrightarrow \quad \frac{dy}{g(y)} = f(x) \, dx$

Once separated, the ODE is just two independent integrals. That's the whole trick — no need for integrating factors, no need for linear-algebra machinery. It is by far the fastest technique when it applies, so always check separability first.

The given equation

$\frac{dy}{dx} = x y$

Here $f(x) = x$ and $g(y) = y$. Separable.

Step-by-step solution

Step 1 — Separate. $\frac{dy}{y} = x \, dx \qquad (y \ne 0)$

Step 2 — Integrate both sides. $\int \frac{dy}{y} = \int x \, dx$ $\ln|y| = \frac{x^2}{2} + C_1$

Step 3 — Solve for $y$ by exponentiating. $|y| = e^{x^2/2 + C_1} = e^{C_1} \, e^{x^2/2}$

Drop the absolute value by letting $C = \pm e^{C_1}$ absorb the sign: $\boxed{\, y(x) = C \, e^{x^2/2} \,}$

Every non-zero value of $C$ gives a different solution curve — that is the general solution, a one-parameter family.

The constant solution we might have lost

When we divided by $y$ in step 1 we implicitly assumed $y \ne 0$. Check $y \equiv 0$ directly in the original ODE: $\frac{d}{dx}(0) = 0, \quad x \cdot 0 = 0 \quad \checkmark$

So $y = 0$ is also a solution — and it corresponds to $C = 0$ in our formula. The general solution $y = C e^{x^2/2}$ covers all solutions, including the zero solution.

Verification

Differentiate $y = C e^{x^2/2}$: $\frac{dy}{dx} = C \cdot x \cdot e^{x^2/2} = x \cdot \bigl(C e^{x^2/2}\bigr) = x y \quad \checkmark$

Initial value problem — picking one curve out of the family

If you're told $y(0) = 3$, plug in: $3 = C e^{0} = C \implies C = 3$

So the unique solution through $(0, 3)$ is $y(x) = 3 \, e^{x^2/2}$

One initial condition ↔ one curve.

Geometric picture

• Every member of the family has the same shape — a Gaussian-like $e^{x^2/2}$ profile — scaled vertically by $C$.
• Positive $C$ gives curves that live above the $x$-axis; negative $C$ mirrors them below.
• The $x$-axis ($C = 0$) is an invariant line — a trajectory that stays put.

When to use separation

Check if the right side factors as $f(x) g(y)$:

• $\dfrac{dy}{dx} = \cos(x) e^{y}$ — separable ($f = \cos x$, $g = e^y$).
• $\dfrac{dy}{dx} = x + y$ — not separable (a sum, not a product).
• $\dfrac{dy}{dx} = \dfrac{x y}{1 + x^2}$ — separable ($f = \frac{x}{1+x^2}$, $g = y$).
• $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$ — not separable, but homogeneous (see #178).

Common mistakes

• Forgetting the absolute value when integrating $1/y$. It matters if you want negative-valued solutions.
• Leaving two constants (one for each integral). There's only one independent constant — fold them into a single $C$.
• Dropping the singular solution $y = 0$ (or more generally wherever $g(y) = 0$) without checking. Always verify it directly.
• Writing $\ln y$ without the absolute value, then being stuck when the data is negative.

Try it in the visualization

Slide the initial condition $y(0)$ to pick a curve out of the family, and watch how all curves share the same Gaussian shape. Toggle the zero solution to see the invariant line $y = 0$.