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Differential equations

  1. Jan 21, 2007 #1
    Here's the problem:
    Solve y'-4y=9e^(7t) with y(0)=5


    multiplying both sides by mu(t)

    now when y(0)=5

    so y=(3e^(3t))/(e^(-4t))+2

    That's not right, I also tried y=(3e^(3t))/(e^(-4t))(5/3)
    and that's not right either. Any idea what I'm doing wrong?


    Never mind, figured it out, I didn't divide c by the left side of the equation.
    Last edited: Jan 21, 2007
  2. jcsd
  3. Jan 23, 2007 #2
    your on the right track. You have the correct integration factor.
    [tex] \mu(t)= e^{-4t} [tex]
  4. Jan 23, 2007 #3
    hmmm, still havent gotten the hang of how to type formulas here. Ok lets if I get it right this time....
    After you set up the problem you actually need to do a full integration on the RHS, and you need to evaluate your LFS at the endpoints of integration...
    [tex] \frac{d}{dt} y(t)e^{-4t}= 9e^{3t} \rightarrow y(t)e^(-4t') \Big^t_0= \int^t_0 9e^{3t'}dt' [\tex]
  5. Jan 23, 2007 #4
    ok sorry , see if i can get it now....
    [tex] \frac{d}{dt} y(t)e^{-4t}= 9e^{3t} \rightarrow y(t)e^(-4t') \Big^t_0= \int^t_0 9e^{3t'}dt' [/tex]
  6. Jan 23, 2007 #5
    last try....
    [tex]\frac{d}{dt} \left[y(t)e^{-4t} \right]= 9e^{3t} \rightarrow y(t)e^{-4t'} \Big| _0^t= \int^t_0 9e^{3t'}dt'[/tex]
    Last edited: Jan 23, 2007
  7. Jan 23, 2007 #6


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    Why do you have primes on the ts? On the right hand side, yes, the t inside the integral is a "dummy" variable and so should be distinguished from the t outside. I would be inclined to use a completely different letter, say s. There certainly shouldn't be a prime on the t in the exponential on the left.
    To finish, what is
    [tex]\int_0^t 9e^{3s}ds[/tex]
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