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Initial Value problem

  1. Jun 19, 2011 #1
    I am having trouble with the below problem:

    y'-(3/2)y= 3t+ 2e^t, y(0)= y0. fine value of y0 that separate solutions that grow positively and negatively as t=> infinity.

    I found p(t) to be -3/2, u(t) to be e^-3t/2
    => e^-3t/2*y' - 3y/2( e^-3t/2)= e^-3t/2(3t+ 2e^t)
    => -2 -4e^t + ce^ 3t/2
    where I found c = y0+6/ e

    Do you guys see any errors in my math so far? i am confused as to find y0 where the solutions diverge (pos. vs neg)

  2. jcsd
  3. Jun 19, 2011 #2


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    I get

    [tex]y=-2t - \frac 4 3 -4e^t+(y_0+\frac{16}{3})e^{\frac 3 2 t}[/tex]

    so you might want to check your arithmetic. Since you have a negative times one exponential and a positive times the other, that might have something to do with the positive vs negative thing.
  4. Jun 20, 2011 #3
    @LC youre right, thanks. I am a little rusty in my integral rules. Got it now
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