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1st order linear ODE

  1. Oct 17, 2006 #1
    Find the solution of the given initial value problem
    y'-y=2te^2t y(0)=1

    so p(t)=-1
    then [tex] \mu (t)= e^/int -1 dt= e^-t [/tex]
    e^-t y'-e^-t y=2te^2te^-t
    d/dt [e^-t y]=2te^t dt
    e^-t =[tex] \int 2te^t dt [/tex]
    e^-t y= 2te^t-2e^t
    y=-2(t-1)+ce^-t
    plugging in the initial conditon gives me
    y(t)=-2(t-1)-e^-t

    Am I doing this right? If not, can someone help?
     
  2. jcsd
  3. Oct 17, 2006 #2
    [tex] I = e^{-\int t} = e^{-t dt} [/tex].

    So you get [tex] ye^{-t} = 2\int te^{t} dt = 2te^{t} - 2e^{t} + C [/tex]

    [tex] y = \frac{2te^{t} - 2e^{t} + C}{e^{-t}} [/tex]

    [tex] y = \frac{2te^{t} - 2e^{t} + 2}{e^{-t}} [/tex]

    [tex] y = 2te^{2t} - 2e^{2t} + 2e^{t} [/tex]
     
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