I found this problem pretty difficult... if anyone could find a method to solving it, it would be great for me in class tomorrow.
Reduce to first order and solve:
y'' + (1+y^-1)*(y')^2=0
thanks
First make the sub $\displaystyle y'=u$
Now take the derivative with respect to x to get
$\displaystyle \frac{d}{dx}y'=\frac{d}{dx}u$
$\displaystyle y''=\frac{du}{dy}\frac{dy}{dx}$ by the chain rule
and note that $\displaystyle \frac{dy}{dx}=y'=u$ so we get
$\displaystyle y''=u\frac{du}{dy}$ so subbing both of these into the equation we get
$\displaystyle u\frac{du}{dy}+(1+y^{-1})u^2=0$
$\displaystyle \frac{du}{u}=-(1+y^{-1})dy$
This should get you started good luck.