Homoegeneous ODEs

Replace $y$ with $yt$ and $x$ with $xt$, if all the $t$s cancel then the ODE is homogeneous.

$$\frac{dy}{dx} = f\left(\frac{y}{x}\right)$$

$$\textrm{let }v = \frac{y}{x} \textrm{ so } y = vx$$

$$\frac{dy}{dx} = \frac{d}{dx}vx$$

$v$ is a function of $x$ so we must use the product rule.

$$\frac{dy}{dx} = v + x\frac{dv}{dx} = f(v)$$

$$\frac{dv}{dx} = \frac{f(v) - v}{x}$$

This is now a separable function, and can be solved as before.

Example: $y(1) = 1$

$$\frac{dy}{dx} = \frac{y}{x^2}(x-y) = \frac{y}{x} - \left(\frac{y}{x}\right)^2$$

$$\textrm{let }v = \frac{y}{x} \textrm{ so } y = vx$$

$$\frac{dy}{dx} = v + x\frac{dv}{dx} = v - v^2 \quad\textrm{ separable}$$

$$\int \frac{-dv}{v^2} = \int \frac{dx}{x}$$

$$\frac{1}{v} = \ln x + c = \frac{x}{y}$$

$$y = \frac{x}{\ln x + c}$$

$$y(1) = 1$$

$$1 = \frac{1}{\ln(1) + c}$$

$$c = 1$$