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2nd order non homogenous DE

  1. Sep 9, 2006 #1
    Does anyone know how to solve this?
    [tex]\frac {d^2V(t)}{dt^2} + \frac{V(t)}{w} = \frac{Vm}{w}[/tex]
  2. jcsd
  3. Sep 9, 2006 #2
    What is the difference between V(t) and plain V, is V just a constant, and are w and m also constants? If so then this should be a relatively simple equation to solve, because you could solve the corresponding homogenous equation and add the constant Vm/w to that general solution and that should give you the general solution to the equation.
  4. Sep 9, 2006 #3
    V(t) is a function of time.w is a constant m is a subscript of another V but they r constant also.If you dont mind pls show me?
  5. Sep 9, 2006 #4


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    You know that this is a "non-homogenous" de so apparently you know something about des. This is pretty close to being a trivial problem!

    If you "try" a solution to the corresponding homogeneous equation,
    [tex]\frac{dV(t)}{dt}+ \frac{V(t)}{w}= 0[/itex]
    of the form V(t)= ert, then V'= rert and V"= r2ert so r2ert+ (1/w)ert= 0 and you get the "characteristic equation" r2+ 1/w= 0. The solutions to that are either [itex]\pm\sqrt{1/w}[/itex] or [itex]\pm i \sqrt{1/w}[/itex] depending on whether w is positive or negative.

    The solutions to the homogenous differential equation are either
    [tex]V(t)= Ce^{\frac{t}{\sqrt{w}}}+ De^{-\frac{t}{\sqrt{w}}}[/tex]
    [tex]V(t)= Ccos(\frac{t}{\sqrt{w}})+ Dsin((\frac{t}{\sqrt{w}})[/tex]
    Again depending on whether w is positive or negative.

    For a "particular solution" to the entire equation, look for V(t)= A, a constant. Then V"(t)= 0 so the equation becomes
    [tex]\frac{A}{w}= \frac{V_m}{w}[/tex]
    so A= V_m.

    If w is positive, the general solution to the entire equation is
    [tex]V(t)= Ce^{\frac{t}{\sqrt{w}}}+ De^{-\frac{t}{\sqrt{w}}}+ V_m[/tex]

    If w is negative, the general solution to the entire equation is
    [tex]V(t)= Ccos(\frac{t}{\sqrt{w}})+ Dsin((\frac{t}{\sqrt{w}})+ V_m[/tex]
  6. Sep 9, 2006 #5
    Thanks for your help...But for particular solution why is it that V(t)=A?What determines that V(t)=A?
  7. Sep 9, 2006 #6
    Take a look at the DE; you want some way for a solution V(t) to give you a constant, so that the non-homogeneity holds...
    In your example, you want the differential equation just to give you a constant, right? So assuming V(t) = A and equating both sides would give you that constant.
    Pretty hard to explain, but let's say you assuming V(t) = at + b; we have:
    0 + (at+b)/w = (Vm/w)
    at + b = Vm
    And, writing it out in a slightly different way:
    at + b = Vm + 0t
    Equating the co-efficients, you have a = 0, Vm = b.
    And that's another way to find the solution.
    But what determines V(t) = A is because you have a V(t) term on the left side and you have a constant on the right side--- and since V'(t) = 0 and V''(t) = 0 for any constant A, V(t) = A would give you the constant as you want it on the right side.
    Hope that made SOME sense. :P I tried.
  8. Sep 9, 2006 #7
    Thanks for all the help...This is a really good forum
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