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Homework Help: Help with integrating differential equation

  1. Jan 6, 2010 #1
    Hi,

    Could someone help me see how the solution of the equation [tex]\frac{1}{\mu} \frac{\partial p}{\partial z} = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho \frac{\partial v}{\partial \rho})[/tex]

    is [tex]v = \frac{1}{4\mu} \frac{\partial p}{\partial z} \rho^{2} + C_{1}ln(\rho) +C_{2}[/tex]

    Thank you

    teeeeee
     
  2. jcsd
  3. Jan 6, 2010 #2
    Hi,

    Could someone help me see how the solution of the equation [tex] \frac{1}{\mu} \frac{\partial p}{\partial z} = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho \frac{\partial v}{\partial \rho}) [/tex] is [tex] v = \frac{1}{4\mu} \frac{\partial p}{\partial z} \rho^{2} + C_{1}ln(\rho) +C_{2} [/tex]

    Thank you

    teeeeee
     
  4. Jan 6, 2010 #3
    Anyone? Please help
     
  5. Jan 6, 2010 #4
    assume partial p over partial z to be some constant 'k'
    expand the right side you now have a second order linear differential equation with a coeff
    something like
    (D^2 + D)v = (k/mu)*rho
    remember rho is not a constant so find the transient solution
    the sum of the steady state and transient solution is your answer
    hope this helped

    ps: i am lazy to use symbols
     
  6. Jan 6, 2010 #5

    LCKurtz

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    It isn't clear what variables depend on what, but you can check this. I will use subscripts for partials. Let's begin by calling

    [tex]w = \frac{1}{\mu} \frac{\partial p}{\partial z} [/tex].


    So you have

    [tex]\rho w = (\rho v_{\rho})_{\rho}[/tex]. Integrate with respect to [itex]\rho[/itex]:

    [tex]\frac {\rho^2} 2 w + C =\rho v_{\rho},\ v_{\rho}=\frac {\rho} 2 w + \frac C {\rho}[/tex]

    Integrate again:

    [tex]v =\frac {\rho^2} 2 w + C\ln \rho + D = \frac {\rho^2} 2\frac{1}{\mu} \frac{\partial p}{\partial z}+ C\ln \rho + D[/tex]

    You can check it. Something is off by a factor of 1/2.
     
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