Solving separable differential equations

A solution of a differential equation is a function that makes the equation true.

You are not looking for a number. You are looking for a formula for $y$ in terms of $x$.

You can always check a candidate by differentiating it. Take the equation

$\frac{dy}{dx} = 2xy$

and try $y = e^{x^2}$. Differentiating, using the chain rule with $u = x^2$:

$\frac{dy}{dx} = 2x\,e^{x^2}$

That is exactly $2xy$, because $y = e^{x^2}$. So $y = e^{x^2}$ is a solution.

Now try $y = 5e^{x^2}$. The 5 is a constant, so it comes along for the ride and the derivative is still $2x \cdot 5e^{x^2} = 2xy$.

So $y = 5e^{x^2}$ is a solution too.

Every constant multiple works, so the answer is a whole family of curves:

$y = Ae^{x^2}$

$A$ is called the arbitrary constant. Different values of $A$ give different curves, and each curve is a solution.

A formula like this, with the constant still in it, is the general solution.

A first-order differential equation is separable when the right-hand side can be written as a product: something involving only $x$, times something involving only $y$.

$\frac{dy}{dx} = f(x)\,g(y)$

For example, $\frac{dy}{dx} = 2xy$ is separable, with $f(x) = 2x$ and $g(y) = y$.

An equation is not separable when the right-hand side cannot be rearranged into (function of $x$) × (function of $y$). Some sums can: $x + xy = x(1+y)$ and $e^{x+y} = e^xe^y$ are separable. Others, like $x + y$, cannot.

For example $\frac{dy}{dx} = x + y$ is not separable. Questions in this topic are always separable; the skill is spotting the product.

1. Rearrange so that $\frac{dy}{dx}$ is alone on the left.
2. Divide both sides by $g(y)$, the whole part containing $y$: $\frac{1}{g(y)}\frac{dy}{dx} = f(x)$.
3. Move $dx$ to the right, so every $y$ is on the left and every $x$ is on the right: $\frac{1}{g(y)}\,dy = f(x)\,dx$.
4. Integrate both sides: $\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$.
5. Add one constant $C$ and tidy up. Solve for $y$ if the question asks for it in the form $y = \ldots$.

Why step 4 is allowed. Let $G(y)$ be an antiderivative of $\frac{1}{g(y)}$ (a function whose derivative is $\frac{1}{g(y)}$), so that $G'(y) = \frac{1}{g(y)}$. By the chain rule, differentiating $G(y)$ with respect to $x$ gives

$\frac{d}{dx}G(y) = \frac{1}{g(y)}\,\frac{dy}{dx}$

and step 2 says this equals $f(x)$. So $G(y)$ is an antiderivative of $f(x)$, which is exactly what $\int f(x)\,dx$ means.

Treating $\frac{dy}{dx}\,dx$ as $dy$ is a reliable shortcut for this chain-rule argument, and Cambridge accepts it.