Differential Equations. Reduction of Order. Calc IV.

1. The problem statement, all variables and given/known data

(x²+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0

2. Relevant equations

reduction of order, and exact equations..

3. The attempt at a solution

let y'=p and y''=p'
Giving: (x²+2p)p'+2xp Let M=2xp and N=(x²+2p)
So, M_p=2x and N_x=2x Therefore, they are exact.
Let f_x=2xp
so, f(x,p)=x²p+h(p)
f_p=x²=h'(p)
comparing to f_p=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))

 

I'm confused, you're substituting back for p before solving for p? Also, y'(0)≠0 y'(0)=1.

Does that mean my equation should be:
y'²+x²y'=1

And from there solve for y?

 

Wow. I got it mixed up with the problem out of the book. It's been a long day. You're right..

So all I have to do is:

y=∫0dx=c

y=∫(-x²)dx=(-x³)/3+c

Therefore, y=∫(-x²)dx=(-x³)/3+c is the answer?

Thanks for helping me by the way.

 

Oh, so now I say because y(0)=1 0/3+c₂=1 so, c₂=1?

Therefore, y=-x³/3+1

What do you mean don't use the same constant? Are you referring to y=∫0dx=c₁?

What do I do with the y=∫0dx=c₁ anyway?

 

Ya, I figured that out after asking. But how do I check that both -x³/3+1 and 1 work?