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Math Help - Integrating Factor.. question...

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

    I have a question about the integrating factor..

    <br />
\frac{{dx}}<br />
{{dt}} + 2x = 2 + 4t - 2e^{ - t} <br />

    <br />
\begin{gathered}<br />
  I.F. \hfill \\<br />
  e^{2t}  \hfill \\ <br />
\end{gathered} <br />

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

    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...

    <br />
e^{2t} x = 4\int {te^{2t}  - 2e^t  + e^{2t} dt} <br />

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

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

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

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