Solution of nonlinear ODE: $x= yy'-(y')^2$

Question

How to solve $x= yy'-(y')^2.$ Can somebody please hint at some substitution or refer any text related to these type of ode.

Answer 1

Start by rearranging your equation: Add both sides by $(y')^2$, then divide both sides by $y'$. We obtain: $$y=\frac{x}{y'}+y'\tag{1}$$ Notice that $(1)$ is a d'Alembert equation. This is because it is in the form: $$y=x\cdot f(y')+g(y')$$ Where $f$ and $g$ are functions of $y'$. These are typically solved by differentiating both sides with respect to $x$: $$y'=\frac{y'-xy''}{(y')^2}+y''$$ This can be rearranged to give: $$y''=\frac{y'\left((y')^2-1\right)}{(y')^2-x}$$ Now substitute $v=\dfrac{dy}{dx}$. We know that $\dfrac{dv(x)}{dx}=\dfrac{1}{\frac{dx(v)}{dv}}$. As a result, we obtain a first-order linear ODE with $v$ as the independent variable and $x$ as the dependent variable. $$\frac{1}{\frac{dx(v)}{dv}}=\frac{v(v^2-1)}{v^2-x} \implies \frac{dx}{dv}+\frac{1}{v(v^2-1)}\cdot x=\frac{v}{v^2-1} \tag{2}$$ This can be solved using an integrating factor.