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?
Best Answer
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 $$ y'(x)=-y(x)-1\implies y(x)=e^{1-x}-1. $$