I have the following equation $y''(x) = y(x) \cdot y'(x) + (y'(x))^2$ with the initial values $y(1) = 0$ and $y'(1) = -1$.

I am seeking some guidance for how to best tackle this particular problem.

this article suggests that a v substitution may reduce the problem in some capacity. Thus I let $v = y'(x)$ as follows:

\begin{align*} y''(x) &= y(x) \cdot y'(x) + (y'(x))^2 & (1) \\ y''(x) &= y(x) \cdot v + v^2 & (2) \\ \end{align*}

When I enter (2) into wolfram, the engine says that the equation is a second-order linear ordinary differential equation.

Can the substitution be leveraged from here to solve the problem? Or is there a better way of doing this?

  • 1
    $\begingroup$ I don't think the solution to this ODE has a closed form. $\endgroup$ – Peter Foreman Mar 27 '19 at 20:01

The idea that is proposed is to try to find an equation for a function $v$ so that $y'(x)=v(y(x))$. By the chain rule, $y''=v'v$ and thus the equation reads as $$ vv'=yv+v^2\implies v=0\lor v'=y+v. $$ The first case gives constant solutions, but contradicts the initicial conditions. In the second case, $v(y)=Ce^y-y-1$. With $v(0)=-1$ one finds $C=0$. For this exceptional case one can now also solve the resulting first order equation $$ y'(x)=-y(x)-1\implies y(x)=e^{1-x}-1. $$


I think Wolfram fails to realize $y$ and $v$ are related. You should end up with the system of equations $$ \left\{ \begin{matrix} v &= y' &\\ v' &= yv &+ v^2 \end{matrix} \right\| $$


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