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Math Help - separable differential equation

  1. #1
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    separable differential equation

    So I've attempted this problem sooo many times! And I don't know where I'm going wrong. I've also consulted many friends in my class and they're also stuck. So here it goes....

    du/dt = e^(2.6t-3.1u), u(0)= 1.9
    Find u(t).

    So what I did was:
    du/dt = e^(2.6t)/e^(3.1u)
    e^(3.1u) du = e^(2.6t) dt
    INTEGRATE BOTH SIDES
    1/3.1 (e^3.1u) = 1/2.6 (e^2.6t) + C

    and then I solved for C:
    C= (1/3.1)e^(5.89) - (1/2.6)

    and then my u(t) equation was:
    [(2.6t(ln(1/2.6)) + 5.89(ln(1/3.1)) - ln (1/2.6))/(ln(1/3.1))]/(3.1)

    which is wrong.
    so if anyone can help out that would be great!!
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  2. #2
    Senior Member Peritus's Avatar
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    \begin{gathered}<br />
  \frac{{du}}<br />
{{dt}} = e^{2.6t - 3.1u}  \hfill \\<br />
  e^{3.1u} du = e^{2.6t} dt \hfill \\<br />
  \frac{1}<br />
{{3.1}}e^{3.1u}  = \frac{1}<br />
{{2.6}}e^{2.6t}  + C \hfill \\ <br />
\end{gathered} <br />

    now apply the initial condition...
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  3. #3
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    yes, I did that.
    And I get C= (1/3.1)*e^(5.89) - (1/2.6)

    is that correct???
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  4. #4
    Senior Member Peritus's Avatar
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    correct
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