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Thread: Integrating Factor.. question...

  1. #1
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    Integrating Factor.. question...

    I have a question about the integrating factor..

    $\displaystyle
    \frac{{dx}}
    {{dt}} + 2x = 2 + 4t - 2e^{ - t}
    $

    $\displaystyle
    \begin{gathered}
    I.F. \hfill \\
    e^{2t} \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    e^{2t} (\frac{{dx}}
    {{dt}} + 2x) = 2e^{2t} + 4te^{2t} - 2e^t \hfill \\
    \frac{d}
    {{dt}}(e^{2t} x) = 2e^{2t} + 4te^{2t} - 2et \hfill \\
    e^{2t} x = e^{2t} + 4\int {te^{2t} dt - 2e^t } \hfill \\
    \end{gathered}
    $

    so I dont get that last part.. my notes tell me that last part is right but shouldnt you integrate the whole expression like this...

    $\displaystyle
    e^{2t} x = 4\int {te^{2t} - 2e^t + e^{2t} dt}
    $

    Which way is right and why...?
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  2. #2
    Senior Member TheAbstractionist's Avatar
    Joined
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    Hi dankelly07.

    $\displaystyle \int\left(2e^{2t} + 4 te^{2t} - 2e^t\right)\,dt$

    $\displaystyle =\ \ \int2e^{2t}\,dt\,+\,\int4 te^{2t}\,dt\,+\,\int- 2e^t\,dt$

    $\displaystyle =\ \ e^{2t}+\underbrace{4\int te^{2t}\,dt}_{\begin{tabular}{c}integrate this\\by parts\end{tabular}}-\ 2e^t$
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