1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Variation of Parameters

  1. Jun 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Given that y=x^2 is a solution to the differential equation:
    (x^2)y'' + 2xy' - 6y = 0 <--- Eq.(1)
    find the general solution of the differential equation
    (x^2)y'' + 2xy' - 6y = 10(x^7) + 15(x^2) <--- Eq.(2)
    Hence write down a second linear dependent solution of equation (1) and a particular solution of equation (2).

    2. Relevant equations
    I've basically concluded that variation of parameters is necessary. I don't think I completely understand what is being asked.

    3. The attempt at a solution
    I tried letting y= V(y1) = V(X^2)
    hence y'= (x^2)V' + 2xV and
    y''= (x^2)V'' + 2xV' + 2V
    Here is where I think I'm getting confused, sub. back in to (1) I get:
    which equals V''(x^4)+4(x^3)V"=0
    This is where I come to a dead end, any help or advice would be greatly appreciated, thank you.
  2. jcsd
  3. Jun 20, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I assume you meant V'' x^4 + 4 x^3 V' = 0? Doesn't that simplify into a form that's very easy to solve?

    Anyways, aren't you trying to solve equation (2)? You should be substituting into that.
  4. Jun 20, 2007 #3
    isn't there another variation of parameters that can be used to find another linearly independent solution of a DE where you take the solution you find from the characterstic, multiplying it by u(x), then differentiating it and plugging it into and solving for u(x)?
  5. Jun 20, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's what he did, except he called it V instead of u(x).
  6. Jun 20, 2007 #5
    Ok so I'm at V''(x^4) + 4(x^3)V'=10(x^7) + 15(x^2). I then just devided all by (x^4) to get V'' alone, hence"
    V'' + V'(4/x) = 10(x^3) + 15/(x^2)
    I then let u=V' and u'=V'', therefore:
    u' +u(4/x) = 10(x^3) + 15/(x^2), I then let P(x)=4/x and hence I(x)=x^4 after integration.
    Now I have (d/dx) (x^4)u = integral of [10(x^7) + 15(x^2).dx],
    hence (x^4)u= (10x^8 / 8) + 5(x^3) + C,
    V'= (5x^4 / 4) + 5/x + C/(x^4) and finally
    V= (x^5)/4 + 5log(x) - C/(3x^3) +D.

    This is wrong and I just don't know where I went wrong. Thanks for the help so far guys.
    Last edited: Jun 20, 2007
  7. Jun 20, 2007 #6
    yea but theres also the use of the wronskian which is called variation of parameters
  8. Jun 20, 2007 #7
    Would anyone know where I went wrong above?
  9. Jun 21, 2007 #8
    Please, anyone?
  10. Jun 21, 2007 #9
    write it out in tex and i'll help but i really can't decypher what you've written
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook