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Solve differential equation

  1. Jan 28, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s1.ipicture.ru/uploads/20120128/U1SBMa6O.jpg

    The attempt at a solution
    I expanded the equation and got:
    [tex]\frac{dy}{dx}=\frac{ylny-ylnx+y}{x}[/tex]
    I can't use the method of separation of variables, nor is this a homogeneous equation. It's not a 1st order linear ODE nor in the Bernoulli format. I'm stuck.
     
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  3. Jan 28, 2012 #2

    ehild

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    Hi sharks,
    The right side of the equation is function of y/x. Try new variable y/x=u.

    ehild
     
    Last edited: Jan 28, 2012
  4. Jan 28, 2012 #3

    sharks

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    Hi ehild!

    So, i guess this is a homogeneous equation.
    [tex]\frac{dy}{dx}=u(lny-lnx+1)[/tex]
    I'm supposed to eliminate all y and x on the R.H.S. and replace by u only.
    [tex]\frac{dy}{dx}=u(lny-lnx+1)=u(lnu+1)=ulnu+u[/tex]
    From [itex]y=ux[/itex], [itex]\frac{dy}{dx}=u+x\frac{du}{dx}[/itex]. Then, [itex]u+x\frac{du}{dx}=ulnu+u[/itex].

    So, [itex]x\frac{du}{dx}=ulnu[/itex]
    [tex]\frac{dx}{x}=\frac{1}{ulnu}\,.du[/tex]
     
    Last edited: Jan 28, 2012
  5. Jan 28, 2012 #4

    ehild

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    And what is dy/dx in terms of u and x?

    ehild
     
  6. Jan 28, 2012 #5

    sharks

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    I have edited my previous post but then i'm stuck at integrating the R.H.S. of: [tex]\int \frac{dx}{x}=\int \frac{1}{ulnu}\,.du[/tex]
    [tex]lnx=\int \frac{1}{ulnu}\,.du[/tex]
    If i consider integrating [itex]1/u[/itex] then i get [itex]lnu[/itex].
     
    Last edited: Jan 28, 2012
  7. Jan 28, 2012 #6

    ehild

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    1/u is the differential of ln(u). So the integrand is of form 1/ln(u) (lnu)'
    I just can not understand what you did.

    ehild
     
  8. Jan 28, 2012 #7

    sharks

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    The integration result would then be [itex]ln(lnu)+lnA[/itex], where lnA is the arbitrary real constant of integration.
    Therefore, [itex]lnx=ln(lnu)+lnA[/itex] or [itex]x=Alnu[/itex] or [itex]\frac{x}{A}=lnu[/itex]
    [tex]\frac{x}{A}=ln\frac{y}{x}[/tex]
    [tex]\huge e^\frac{x}{A}=\frac{y}{x}[/tex]
    So, the final answer is (i'm using big font to make the power fraction more visible):
    [tex]\huge y=xe^\frac{x}{A}[/tex]
    Is that correct?
     
    Last edited: Jan 28, 2012
  9. Jan 28, 2012 #8

    ehild

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    Try to substitute it back into the original equation.

    ehild
     
  10. Jan 28, 2012 #9

    sharks

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    Yes, i get the same L.H.S and R.H.S. that is,
    [tex]\huge \frac{xe^\frac{x}{A}}{A}+e^\frac{x}{A}[/tex]
    So, i assume this is the proof. Thanks, ehild.:smile:
     
  11. Jan 28, 2012 #10

    ehild

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    You are welcome:smile: Remember that y/x substitution.

    ehild
     
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