# Reduction to first order

• Sep 30th 2009, 05:08 PM
ntfabolous
Reduction to first order
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
• Sep 30th 2009, 05:45 PM
TheEmptySet
Quote:

Originally Posted by ntfabolous
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.