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How does one solve this differential equation?

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the general solutions to [tex] \frac {dx}{dt} = \frac {x+t}{x-t} [/tex]
    Edit: Sorry I messed up the original equation.

    2. Relevant equations
    No idea.


    3. The attempt at a solution
    I tried to use substitution u=x+t to change it into a separable equation but I only got to:
    [tex] \frac {du}{dt} = \frac {2u-2t}{u-2t}[/tex]
    which I don't think is separable.
     
    Last edited: Oct 2, 2011
  2. jcsd
  3. Oct 2, 2011 #2
    Actually, the original differential equation is separable. Try separating and integrating accordingly. After all, your right hand side is a function only of x.
     
  4. Oct 2, 2011 #3
    Sorry I messed up the original equation. The one I need help on isn't as simple. :/
     
  5. Oct 2, 2011 #4
    Actually I just solved this question after much head scratching. I'm going to post it here for reference if needed because I know a lot of Google searches point to Physics Forums.

    Letting [tex] u=\frac{x}{t}[/tex] then [tex]\frac{dx}{dt}=t\frac{du}{dt}+u[/tex]
    Substituting into original DE you'll get [tex]t\frac{du}{dt} = \frac{-u^2+2u+1}{u-1}[/tex]
    This then becomes separable and simply solve for [tex]\int\frac{u-1}{-u^2+2u+1}du=\int\frac{1}{t}dt[/tex] and make sure to back-substitute.

    I'm pretty sure this is right but I'm not 100%. I still do not understand how do we choose the substitution equation for DE or is it similar to choosing substitution for integration?
     
  6. Oct 2, 2011 #5

    ehild

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    Homework Helper
    Gold Member

    It is right.

    Your original equation can be written in the form [itex]x'=\frac{1+x/t}{1-x/t}[/itex]

    When the DE is the form of y'=F(y/x) substitute y/x =u.

    Then y=ux, y'=u'+u, so u'+u=F(u), u'=F(u)-u, a separable equation.

    See this place: http://www.math.hmc.edu/calculus/tutorials/odes/odes.pdf

    ehild
     
    Last edited: Oct 2, 2011
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