# Differential Equations. Reduction of Order. Calc IV.

Gummy Bear

## 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))

Homework Helper
Gold Member

## 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

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.

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?

Homework Helper
Gold Member

## Homework Statement

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

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.

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

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

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.

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.

Homework Helper
Gold Member
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.

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.

Homework Helper
Gold Member
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.

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?

Homework Helper
Gold Member
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?

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.

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