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Second Order Nonhomogeneous

  1. Oct 18, 2005 #1
    The following equation was derived from a RLC circuit:

    [tex]\frac{d^2}{dt^2} (V(t)) + 6 \frac{d}{dt} (V(t)) + 5V(t) = 40[/tex]

    Setting up the equation:

    [tex]s^2 +6s + 5 = 0[/tex]

    yields [tex]s = -1[/tex] and [tex]s = -5[/tex]

    Giving me the general equation:

    [tex]V(t) = k_{1}e^{-t} + k_{2}e^{-5t}[/tex]

    But the general equation shown in the solution is:

    [tex]V(t) = k_{1}e^{-t} + k_{2}e^{-5t} + k_{3}[/tex]

    It's been a little while since my differential equations class and I'm not sure where the k3 comes from. Is it because the equation is nonhomogeneous and if that's the case will all second order nonhomogeneous equations that equal a constant have similar general equations?
    Last edited: Oct 18, 2005
  2. jcsd
  3. Oct 18, 2005 #2


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    Yes it is because is nonhomogenous, the particular solution for this equation will have been 8. [itex] v(t)_{p} = 8 [/itex]
    Last edited: Oct 19, 2005
  4. Oct 19, 2005 #3


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    Shouldn't [itex] k_{3}=8 [/itex] ??

  5. Oct 19, 2005 #4
    Okay guys I got it, thanks for your help. And yes [tex]k_{3} = 8[/tex]
  6. Oct 19, 2005 #5


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    [tex]V(t) = k_{1}e^{-t} + k_{2}e^{-5t} + k_{3}[/tex]

    In my opinion, that is really bad notation! The fact that the same symbol is used for [itex]k_1, k_2, k_3[/itex] implies that they are to be treated the same. In fact, k1 and k2 are arbitrary constants while k3 must be 8.
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