Solve x^2 (dy/dx)^2-2xy dy/dx+2y^2-x^2=0 | Equation Solvable for p
Equation Solvable for p MathJax Example Solve \(x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0\). Solution: \[ \begin{align*} & \text{The given equation is} \\ & x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{2xy \pm \sqrt{4x^2 y^2 - 4x^2 (2y^2 - x^2)}}{2x^2} \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{y \pm \sqrt{x^2 - y^2}}{x} \\ \end{align*} \] Put \(y = vx\) so that \(\frac{dy}{dx} = v + x \frac{dv}{dx}\). Therefore, (2) becomes \(v + x \frac{dv}{dx} = \frac{vx \pm \sqrt{x^2 - v^2 x^2}}{x}\). Therefore, \(v + x \frac{dv}{dx} = v \pm \sqrt{1 - v^2}\). Therefore, \(x \frac{dv}{dx} = \pm \sqrt{1 - v^2}\). Separating the variables, \(\frac{1}{\sqrt{1 - v^2}} dv = \pm \frac{1}{x} dx\). Ma...