Integrating Factor.. question...

**dankelly07** · May 21st 2009, 04:21 AM

I have a question about the integrating factor..

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

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

$ \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...

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

Which way is right and why...?

**TheAbstractionist** · May 21st 2009, 04:43 AM

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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