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

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    $\begingroup$ it is a D' Alembert equation, Google for it $\endgroup$ – Dr. Sonnhard Graubner Jun 29 '17 at 13:44

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.

  • $\begingroup$ I got the solution in terms of $ x (v)$.Now how to proceed. $x (v)=\frac {vlogc (v+\sqrt (v^2-1))}{\sqrt (v^2-1)}$ $\endgroup$ – Upstart Jun 29 '17 at 15:45
  • $\begingroup$ Using the quadratic formula on your original ODE: $$y'=\frac{y\pm \sqrt{y^2-4x}}{2} \tag{3}$$ Substitute back $v=y'$ on your solution for $x(v)$, and substitute $(3)$ to obtain $x$ as a function of $y$. $\endgroup$ – projectilemotion Jun 29 '17 at 15:49

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