Separable ODE

Definition

$g(y)y^{\prime }=f(x)$

$g(y)dy=f(x)dx$

Solution

can be solved by integration $\int g(y)dy=\int f(x)dx+C$

Reduction to Separable ODE

If there is an ODE of the form $y^{\prime }=f(\frac{y}{x})$, then let $\frac{y}{x}=u$. Then $y=xu\Rightarrow y^{\prime }=u^{\prime }x+u\Rightarrow u^{\prime }x+u=f(u)\Rightarrow \frac{du}{f(u)-u}=\frac{dx}{x}$:

Examples

&\cfrac{dy}{dx} = (y^{2} + 1)\\ &\cfrac{1}{y^{2}+1}dy = dx\\ &\int \cfrac{1}{y^{2}+1}dy = \int 1 dx = tan^{-1}(y) = x+C\\ &y = \tan(x+C) \end{aligned}$$ $$\begin{aligned} &2xy\frac{dy}{dx} = y^{2}-x^{2}\\ & \frac{dy}{dx} = \frac{1}{2}\left( \frac{y}{x} - \frac{x}{y} \right)\\ &\text{substitute}\ u:=\frac{y}{x} \Rightarrow u+x\frac{du}{dx} = \frac{dy}{dx}\\ & u+x\frac{du}{dx} = \frac{1}{2}\left( \frac{u^{2}-1}{u} \right)\\ & x\frac{du}{dx} = -\frac{1}{2}\left( \frac{u^{2}+1}{u} \right)\\ & \int\left( \frac{2u}{u^{2}+1} \right)du = -\int\frac{1}{x}dx\\ & \ln|u^{2}+1| = -\ln|x|+C\\ & |u^{2}+1| = \frac{1}{x}e^{C}\\ & u^{2}+1 = \pm \frac{1}{x}e^{C}\\ & \frac{y^{2}}{x^{2}} = \frac{1}{x}C -1\\ & y = \pm \sqrt{Cx - x^{2}} \end{aligned}$$