# Help with solving this differential equation?

1. Sep 18, 2012

### lillybeans

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

(3y4-11x2y2-28x4)dx-(4xy3)dy=0

2. Relevant equations

My=Nx if equation is exact, if not, I can make it exact (I hope) by finding an integrating factor. The problem is I can't get the integrating factor to be a function of x or y only.

3. The attempt at a solution

Let stuff in front of dx=M
Let stuff in front of dy=N

My=12y3-22x2y
Nx=-4y3
My-Nx=16y3-22x2y

First try (My-Nx)/N=(16y3-22x2y)/4xy3

Not a function of x only.

Then (My-Nx)/-M=16y3-22x2y/-(3y4-11x2y2)

Not a function of y only.

Am I using the wrong method to solve this question? I don't see any other way to go about it rather than turning this into an exact equation. Why can't I find an integrating factor that has a pure x or y expression?

Thanks.

2. Sep 18, 2012

### LCKurtz

M(x,y) and N(x,y) are both homogeneous of degree 4. So y = ux will make a separable equation of it.

3. Sep 18, 2012

### lillybeans

Sorry, I haven't quite learned that yet. Do you mind elaborating a little? Is that another technique for solving differential equations? Something called substitution, maybe?

Thank you

4. Sep 18, 2012

### LCKurtz

A function $f(x,y)$ is homogeneous of degree $n$ if $f(\lambda x,\lambda y) =\lambda^nf(x,y)$. What $n$ works for your $M$ and $N$?

Try the substitution $y = ux,\quad dy =xdu + udx$.

[Edit, added later]: I see I forgot to tell what this type of DE is called. If is called a first order homogeneous equation. Unfortunately, this is not the same meaning of the term as when applied to a linear DE with 0 on the right side, where we refer to homogeneous vs. non-homogeneous equations. In the setting of your problem, it refers to the $M$ and $N$ being homogeneous in the above sense.

Last edited: Sep 18, 2012
5. Sep 18, 2012

### lillybeans

[solved]

Last edited: Sep 18, 2012
6. Sep 18, 2012

### lillybeans

Nevermind I made a mistake, that's why I thought I couldn't turn the last one into an exact one.

Thank you once again! :)