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Simple linear differential equation

  1. Apr 13, 2010 #1
    Hey, I know this is easy I just can't remember how to do it.


    and [tex]By''+y'=A [/tex] A,B constants.

    So complementary solution [tex] Bm^{2}+m=0 \\ \text{ therefore } m=0, \frac{-1}{B}[/tex]
    therfore [tex] y_{C} = C_{1}+C_{2}e^{-t/B} [/tex]
    Not sure what to do for particular solution though because substituting in constant doesn't give you anything

    The final answer is:

    [tex] y= A[t+B(e^{-t/B}-1] [/tex]

  2. jcsd
  3. Apr 13, 2010 #2


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    Since y = constant is a solution for the homogeneous equation, it can't be a solution yp of the NH equation. For your particular solution try yp = Ct.
  4. Apr 14, 2010 #3
    Ah yes now I remember! Thank you very much
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