Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homemogeneous Equation

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of each homogeneous equation.

    2. Relevant equations


    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. jcsd
  3. Aug 9, 2011 #2
    to integrate dv/1-2v^2+v, factor the bottom and use partial fractions.
  4. Aug 10, 2011 #3
    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,,,
  5. Aug 10, 2011 #4
    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.
  6. Aug 10, 2011 #5
    Oh...I've never heard about that...but thank you so much!
  7. Aug 10, 2011 #6


    User Avatar
    Homework Helper

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

  8. Aug 10, 2011 #7
    I actually could figure it out! Thank you so much guys!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook