# Integrating Factor.. question...

• May 21st 2009, 04:21 AM
dankelly07
Integrating Factor.. question...
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...?(Thinking)
• May 21st 2009, 04:43 AM
TheAbstractionist
Hi dankelly07.

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

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

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