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: Nonlinear differential equation

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data

    In one problem I had got to this equations, but I was not able to solve it, because I'm actually
    on high school.

    The equation :
    d^2/dt^2(x) = -h*g/(h+x)

    I tried use separation of variables but I was not able to use the chain rule.
    Can anybody show me the steps with explanation?
    Last edited by a moderator: May 9, 2012
  2. jcsd
  3. May 9, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi mbadin! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    either multiply both sides by dx/dt

    or use d2x/dt2 = vdv/dx

    (where v = dx/dt … you can prove it using the chain rule :wink:)
  4. May 9, 2012 #3


    User Avatar
    Gold Member

    OK, here is how your equation appears to me:
    [tex]\frac{d^2x}{dt^2} = -\frac{hg}{h+x}[/tex]
    If i understood your question correctly, then you need to integrate twice to get rid of the differentiation.
  5. May 9, 2012 #4


    User Avatar
    Science Advisor

    It's not that easy because the right side, that you want to integrate with respect to t depends upon the unknown function x.

    I would do what tiny-tim suggested: Let v= dx/dt so that [itex]d^2x/dt^2= dv/dt[/itex] and then, by the chain rule,
    [tex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/tex]

    Your equation becomes
    [tex]v\frac{dv}{dx}= -\frac{hg}{h+ x}[/tex]
    which can be integrated as
    [tex]\int v dv= hg\int \frac{dx}{h+ x}[/tex]

    It might well give you a function v= dx/dt that is difficult to integrate but that is the most direct method to solve this equation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook