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Finding the general solution of the given differential equation

  1. Dec 13, 2013 #1
    y"+2y'+y=2e^-t
    I tried to find the solution for this nonhomogenous diff. Equation but i could not. First i took a function Y(t)=Ae^-t but i was getting 0=2e^-t.
    To get rid of that i took another y'+y=2e^-t and found the solution y=2te^-t + ce^-t. Noticed that first part of this finding is solution of my nonhomogenous diff equation so i took another function Y(t)=Ate^-t but then again i am finding the same answer, i need some tips on how to continue. The general solution for this equation as homogenous equation has repeated roots.
     
  2. jcsd
  3. Dec 13, 2013 #2

    ShayanJ

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    Gold Member

    When the characteristic equation has repeated roots,the answer is [itex]Ae^{qt}+Bte^{qt} [/itex].
    For finding the particular solution,take [itex] y_p=f(t)e^{-t} [/itex] and try finding f(t).
     
  4. Dec 13, 2013 #3

    CAF123

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    The homogeneous differential equation is second order and so the general solution is composed of two linearly independent pieces. One of those pieces is precisely of the form Y(t) = Ate-t, so subbing this in as a particular solution is redundant. You have to look for another form.
     
  5. Dec 14, 2013 #4
    I multiplied my function by t^2, since by multiplying only by t was not working, and it worked.
     
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