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: Second order ODE

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data

    find the general solution to the ODE:

    2. Relevant equations

    3. The attempt at a solution

    I put this in the following form: y''=-(1/x)y'
    integrated both sides: y'=-ln(x)y +C
    I think i made a stupid mistake but i cant figure out what it is.
    Any help would be appreciated. Thank you.
  2. jcsd
  3. Mar 10, 2009 #2
    You can't integrate like that. You can only integrate when you have an ODE of the form y' = f(x), where anything y isn't on the right side.

    There is a general solution to ODEs of the form y'' + a(x)y' = 0, and it's probably in your textbook. Hint: multiply both sides by [tex]e^{\int^x_{x_0} a(t)dt}[/tex], and contract by the product rule for differentiation.
  4. Mar 10, 2009 #3
    why do i have to multiply both sides by that integral?
  5. Mar 10, 2009 #4
    Because it gives you something you can integrate. You're trying to find y, right? Well, if you multiply both sides by the given function, and note that the derivative of [tex]e^{\int^x_{x_0} a(t)dt}[/tex] is [tex]a(x)e^{\int^x_{x_0} a(t)dt}[/tex], you can use the product rule to write the equation in the form (X y')' = 0, where X is something you need to find yourself (I've given you all the clues you need). You can then conclude that X y' = c, for some constant c, by integrating both sides. You now have an equation that is easily solvable if you know the very basics of differential equations.
  6. Mar 12, 2009 #5
    ok, so i did this and my working out is:
    so the 'a' is 1/x, and integrating this gives lnx so e^ln(x)=x so multiplying the ODE vy x gives:
    so (xY')'=0
    integrating both sides gives:
    xY'=C (where C is a constant)
    integrating again gives:
    Y=Cln(x)+D (where D is a constant)
    Is this correct?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook