General Solution for Homogeneous Equations: (x^2)y'=2(y^2)-x^2

[peace-Econ]

Homework Statement

Find the general solution of each homogeneous equation.

Homework Equations

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

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.

 

[cragar]

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

 

[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,,,

 

[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.

 

[peace-Econ]

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

 

[ehild]

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,,,

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

ehild

 

[peace-Econ]

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