(x2+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0
reduction of order, and exact equations..
The Attempt at a Solution
let y'=p and y''=p'
Giving: (x2+2p)p'+2xp Let M=2xp and N=(x2+2p)
So, Mp=2x and Nx=2x Therefore, they are exact.
comparing to fp=x2+2p we know h'(p)=2p so, h(p)=p2
Now, solve for p: using quadratic.
Substitute back for y'=p
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: