# Homemogeneous Equation

1. Aug 9, 2011

### peace-Econ

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

Find the general solution of each homogeneous equation.

2. Relevant equations

(x^2)y'=2(y^2)-x^2

3. The attempt at a solution

Because y'=(dy/dx), I changed the equation to (x^2-2y^2)dx+x^2dy=0
Homogeneous of degree is 2.

I let y=xv, dy=vdx+xdv
So, I have (x^2-2x^2v^2)dx+x^2(vdx+xdv)=0
This equals to (1-2v^2+v)dx+xdv=0

Then, the integrating factor is 1/x(1-2v^2+v)

so, dx/x+dv/1-2^2+v=0

Here, I need to integrate dv/1-2v^2+v, but I don't how to do it.
So, does anyone help me calculate it? or if you find any mistake in my work, please please let me know.

Thank you so much.

2. Aug 9, 2011

### cragar

to integrate dv/1-2v^2+v, factor the bottom and use partial fractions.

3. Aug 10, 2011

### peace-Econ

You're right. I totally forget it. Thank you so much!

Sorry, can you help me one moe thing?
How can i factor x^3-2x-1? This is kinda killing me now,,,

4. Aug 10, 2011

### cragar

you need to use synthetic division. look at the factor of p/q where p=-1 and q= 1 these numbers come from the coefficients of the x^3 term and the constants term from the polynomial. then once you find a factor and you have done synthetic division you will now have a polynomial of degree 2 which you can factor again.

5. Aug 10, 2011

### peace-Econ

Oh...I've never heard about that...but thank you so much!

6. Aug 10, 2011

### ehild

Note that x^3-2x-1= (x^3-x)-(x+1)

ehild

7. Aug 10, 2011

### peace-Econ

I actually could figure it out! Thank you so much guys!