# Differential Equations. Reduction of Order. Calc IV.

1. Sep 13, 2012

### Gummy Bear

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, 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))

2. Sep 13, 2012

### LCKurtz

Right there you have $y'^2+x^2y'=c$. If you use $y'(0)=0$ you get $c=0$ so your equation is $y'^2+x^2y'=y'(y'+x^2)=0$. This tells you that either $y'=0$ or $y'=-x^2$, both of which are easy to handle.

3. Sep 13, 2012

### Gummy Bear

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?

4. Sep 13, 2012

### LCKurtz

I have quoted your original boundary condition. $p=y'$ so I just put it in. It doesn't matter which you call it, but using $y'$ emphasizes that is what it is, and it makes it more natural when using the boundary conditions.

No. You get 0 on the right side. It means either $y'=0$ or $y'=-x^2$. Solve those, apply any boundary conditions and, if you aren't sure, check the solutions work.

5. Sep 13, 2012

### Gummy Bear

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

Thanks for helping me by the way.

6. Sep 13, 2012

### LCKurtz

Use the boundary condition to evaluate c. A second order equation with two initial conditions won't have any arbitrary constants left. And you shouldn't use the same c for both constants.

7. Sep 13, 2012

### LCKurtz

I should add here that this is a non-linear equation. You get two separate solutions. Nothing says that the sum of those two solutions is also a solution.

8. Sep 13, 2012

### Gummy Bear

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?

9. Sep 13, 2012

### LCKurtz

You apply the boundary condition $y(0)=1$. Remember you used $y'(0)=0$ to get the two separate solutions. Each of them satisfies $y(0)=1$. So you have two different solutions to this problem. You should check that they both work.

10. Sep 14, 2012

### Gummy Bear

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

11. Sep 14, 2012

### LCKurtz

Plug them into the equation and boundary conditions.