- #1
Gummy Bear
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Homework Statement
(x2+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: (x2+2p)p'+2xp Let M=2xp and N=(x2+2p)
So, Mp=2x and Nx=2x Therefore, they are exact.
Let fx=2xp
so, f(x,p)=x2p+h(p)
fp=x2=h'(p)
comparing to fp=x2+2p we know h'(p)=2p so, h(p)=p2
so, c=p2+px2
Now, solve for p: using quadratic.
p=(-x2±√(x4+4c))/2
Substitute back for y'=p
y'=(1/2)(-x2±√(x4+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)(-x2±√(x4))