Variable-separable equations

The simplest first-order differential equations have the form $\frac{dy}{dx} = f(x)\,g(y).$ The right side is a product of a function of $x$ alone and a function of $y$ alone. Whenever you can write a DE in this form, the solution method is mechanical.

The method

If $g(y) \neq 0$, divide: $\frac{1}{g(y)}\,dy = f(x)\,dx.$ Now each side depends on only one variable. Integrate both sides: $\int \frac{dy}{g(y)} = \int f(x)\,dx + C.$ This is an implicit relation between $x$ and $y$. Often you can solve for $y$ explicitly; sometimes you leave the answer implicit.

If $g(y_0) = 0$ for some constant $y_0$, then $y = y_0$ is also a solution , a constant solution. Check these separately; they may or may not be part of the general family captured by the integration above.

Why does this work?

The chain rule. If $y = y(x)$ satisfies the equation, then $\frac{d}{dx}\int \frac{dy}{g(y)} = \frac{1}{g(y)} \frac{dy}{dx} = f(x) = \frac{d}{dx}\int f(x)\,dx,$ so the two sides differ by at most a constant. Hence $\int \frac{dy}{g(y)} - \int f(x)\,dx = C.$ The single constant $C$ absorbs both constants of integration.

Recognition

The form $\dfrac{dy}{dx} = f(x)g(y)$ may be disguised. Sometimes you must rearrange, e.g.

• $\dfrac{dy}{dx} = (x + 3)(y^2 + 1)$: directly separable.
• $\dfrac{dy}{dx} = e^{x+y}$: rewrite as $e^x \cdot e^y$; separable.
• $\dfrac{dy}{dx} + \dfrac{y}{x} = 0$: rewrite as $\dfrac{dy}{dx} = -\dfrac{y}{x}$; separable.
• $(1 + y^2)\,dy = (1 + x^2)\,dx$: already separated.

If you see a sum that can't be factored into $f(x) g(y)$, the equation is not separable. Move on to other techniques.

Initial-value problems

When a condition like $y(x_0) = y_0$ is given, you solve the general implicit relation, then substitute to find $C$. The result is a particular solution.

Worked examples

Example 1. Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$.

Separate: $\dfrac{dy}{y} = \dfrac{dx}{x}$. Integrate: $\ln|y| = \ln|x| + C_1$. So $|y| = e^{C_1} \cdot |x|$, hence $y = Cx$ (with $C \in \mathbb{R}$ absorbing the sign and the exponential). Family of lines through the origin.

Example 2. Solve $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Separate: $y\,dy = -x\,dx$. Integrate: $\dfrac{y^2}{2} = -\dfrac{x^2}{2} + C_1$, so $x^2 + y^2 = C$ (a family of circles centred at the origin).

Example 3. Solve $\dfrac{dy}{dx} = (1 + x^2)(1 + y^2)$.

Separate: $\dfrac{dy}{1 + y^2} = (1 + x^2)\,dx$. Integrate: $\tan^{-1}y = x + \dfrac{x^3}{3} + C$.

Example 4. Solve $\dfrac{dy}{dx} = e^{x + y}$, $y(0) = 0$.

Rewrite: $\dfrac{dy}{dx} = e^x e^y$. Separate: $e^{-y}\,dy = e^x\,dx$. Integrate: $-e^{-y} = e^x + C$. Apply $y(0) = 0$: $-1 = 1 + C$, so $C = -2$. Particular solution: $-e^{-y} = e^x - 2$, or $e^{-y} = 2 - e^x$.

Example 5. Solve $x\,dy - y\,dx = 0$.

Rewrite: $\dfrac{dy}{y} = \dfrac{dx}{x}$. Integrate: $\ln|y| = \ln|x| + C_1$, so $y = Cx$.

Example 6. Newton's law of cooling: $\dfrac{dT}{dt} = -k(T - T_0)$ with $T(0) = T_1$, $T_0$ ambient.

Separate: $\dfrac{dT}{T - T_0} = -k\,dt$. Integrate: $\ln|T - T_0| = -kt + C_1$, so $T - T_0 = A e^{-kt}$. Apply $T(0) = T_1$: $A = T_1 - T_0$. Final: $T(t) = T_0 + (T_1 - T_0)e^{-kt}$.

Try it yourself

1. $\dfrac{dy}{dx} = x y^2$.
2. $\dfrac{dy}{dx} = \dfrac{y}{x^2}$.
3. $(1 + x)\dfrac{dy}{dx} - y = 0$.
4. $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$.
5. $\dfrac{dy}{dx} = e^{x - y}$.
6. $\dfrac{dy}{dx} = \sec^2 x \cdot \tan y$.
7. $(x^2 + 1)\dfrac{dy}{dx} + 1 = 0$.
8. $\dfrac{dy}{dx} = \dfrac{x(1 + y^2)}{y(1 + x^2)}$.
9. $\sin x \cos y\,dx + \cos x \sin y\,dy = 0$.
10. $\dfrac{dy}{dx} = e^x \cdot \dfrac{1}{y}$, $y(0) = 2$.
11. Radioactive decay: $\dfrac{dN}{dt} = -\lambda N$, $N(0) = N_0$.
12. $\dfrac{dy}{dx} = y \cot x$.
13. $(1 + y^2)\,dx + (1 + x^2)\,dy = 0$.
14. $\dfrac{dy}{dx} = \dfrac{x e^x \log x + e^x}{x \cos y}$, $y(1) = 0$.

Pitfalls and tricks

• Factor first. Many DEs look non-separable until you factor.
• Don't forget constant solutions. If $g(y_0) = 0$, then $y \equiv y_0$ may be missed by dividing.
• Logarithm absorbs constants. $\ln|y| = \ln|x| + C$ becomes $y = Cx$ , the new $C$ has absorbed both sign and $e^{C_1}$.
• Apply initial conditions to the implicit form before solving explicitly. Sometimes it's cleaner.
• Watch domains. Where is $g(y) \neq 0$? The solution exists only there.