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Homogeneous and particular solution

  1. May 19, 2005 #1
    hi,

    I have this ODE and I need to obtain the general and the particular solution, this is the ODE

    Vz''+1/r*Vz'= k

    where k is a constant

    thanks
     
  2. jcsd
  3. May 19, 2005 #2

    dextercioby

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    Who's "V"...?

    Daniel.
     
  4. May 19, 2005 #3

    arildno

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    I think it is a "she"..
     
  5. May 19, 2005 #4
    Vz is just the velocity in the z direction, It's nothing more than my variable. Vz changed with respect to r.

    Vz(r)''+1/r*Vz(r)' = k
     
  6. May 19, 2005 #5

    dextercioby

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    A what...?A "she"...?:confused:

    Daniel.
     
  7. May 19, 2005 #6

    dextercioby

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    Make the substitution

    [tex] \frac{dv_{z}}{dr}=u(r) [/tex]

    Daniel.
     
  8. May 19, 2005 #7

    HallsofIvy

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    I THINK (it surely isn't clear) that the OP mean that Vz is the function to be solved for.

    Assuming also that r is the independent variable, we can multiply the entire equation by r2 to get the "Euler-type" (or "equipotential") equation r2Vz"+ rVz'= kr2. If we let x= ln r, (so that r= ex) we can rewrite that as a linear equation with constant coefficients. Specifically, (I'm writing "y" in place of "Vz" just because it is easier) dy/dr= dy/dx dx/dr= (1/r)(dy/dx) and d2y/dr2= d/dr((1/r)dy/dx)= (-1/r2(dy/dx)+ (1/r2)d2y/dx2.
    The equation becomes d2y/dx2= ke2x. Integrating once, dy/dt= (k/2)e2x+ C1. Integrating a second time, y= (k/4)e2x+ C1x+ C2. ex= r so e2x= r2. In terms of r, y(r)= Vz(r)= (k/4)r2+ C1ln r+ C2.
     
    Last edited: May 19, 2005
  9. May 19, 2005 #8

    arildno

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    Perhaps I was wrong then..
     
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