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## Homework Statement

(x

^{2}+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0

## Homework Equations

reduction of order, and exact equations..

## The Attempt at a Solution

let y'=p and y''=p'

Giving: (x

^{2}+2p)p'+2xp Let M=2xp and N=(x

^{2}+2p)

So, M

_{p}=2x and N

_{x}=2x Therefore, they are exact.

Let f

_{x}=2xp

so, f(x,p)=x

^{2}p+h(p)

f

_{p}=x

^{2}=h'(p)

comparing to f

_{p}=x

^{2}+2p we know h'(p)=2p so, h(p)=p

^{2}

so, c=p

^{2}+px

^{2}

Now, solve for p: using quadratic.

p=(-x

^{2}±√(x

^{4}+4c))/2

Substitute back for y'=p

y'=(1/2)(-x

^{2}±√(x

^{4}+4c)

y=?

This is where i'm not sure what to do.

I'm pretty sure I use the initial conditions, do I assume c=0 and take the integral of:

(1/2)(-x

^{2}±√(x

^{4}))