Non-linear ODE $1+y'^2=yy'^2$, $y(0)=b,y(a)=c$

Question

How to solve $$1+y'^2=yy'^2?$$ The initial condition is $y(0)=b,y(a)=c$.

It is a non-linear ODE.

Some hint is also helpful! Thanks for your help!

Answer 1

We have $$1+(y')^2=y(y')^2$$ Divide by $(y')^2$ to get $$\frac{1}{(y')^2}=y-1$$

Let us solve for $x=x(y)$ instead of for $y=y(x)$. Then we get the equation: $$x'=\sqrt{y-1}\ \ \ \ \ \ \text{We need to do the other sign of the square root too.}$$

Separate variables now: $$\begin{align}dx&=\sqrt{y-1}dy\\x+C&=\frac{2}{3}(y-1)^{3/2}\\y&=\left[\frac{3}{2}(x+C)\right]^{2/3}+1\end{align}$$

And we need to check which of these solutions do satisfy the original equation.