Homogenous Equation

Problem:

$ (y^2 - x^2)dx + xydy = 0 $

Solution:

$ \begin{eqnarray*} (y^2 - x^2)dx + xydy &=& 0 \\ (y^2 - x^2)dx &=& -xydy \\ \frac{dy}{dx} &=& \frac{x^2 - y^2}{xy} \\ &=& \frac{x}{y} - \frac{y}{x} \\ \end{eqnarray*} $

Now let $ z = \frac{y}{x}, \frac{dz}{dx} = \frac{d}{dx}(y)\frac{1}{x} + y\frac{d}{dx}\frac{1}{x} = \frac{dy}{dx}\frac{1}{x} - \frac{y}{x^2} $, which implies $ \frac{dy}{dx} = x\frac{dz}{dx} + z $. Putting these back, we have

$ \begin{eqnarray*} \frac{dy}{dx} &=& \frac{x}{y} - \frac{y}{x} \\ x\frac{dz}{dx} + z &=& \frac{1}{z} - z \\ x\frac{dz}{dx} &=& \frac{1}{z} - 2z \\ \frac{zdz}{1-2z^2} &=& \frac{dx}{x} \\ \frac{-1}{4}\log(1-2z^2) &=& \log(x) + C_1 \\ \log(1-2z^2) &=& -4\log(x) + C_2 \\ 1-2z^2 &=& C_3x^{-4} \\ 1-2(\frac{y}{x})^2 &=& C_3x^{-4} \\ x^2-2y^2 &=& C_3x^{-2} \\ y^2 &=& \frac{1}{2}(x^2 - Cx^{-2}) \\ \end{eqnarray*} $

For a quick check, we have:

$ \begin{eqnarray*} y^2 &=& \frac{1}{2}(x^2 - Cx^{-2}) \\ 2y \frac{dy}{dx} &=& x + Cx^{-3} \\ &=& x + \frac{Cx^{-2}}{x} \\ &=& x + \frac{x^2 - 2y^2}{x} \\ &=& 2x - \frac{2y^2}{x} \\ \frac{dy}{dx} &=& \frac{x}{y} - \frac{y}{x} \\ \end{eqnarray*} $