# Solving an initial value problem involving the second-order nonlinear ordinary differential equation $y''(x) = y(x) \cdot y'(x) + (y'(x))^2$

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?

• I don't think the solution to this ODE has a closed form. – 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 $$\left\{ \begin{matrix} v &= y' &\\ v' &= yv &+ v^2 \end{matrix} \right\|$$