solving differential equation using integration factor

**nivek0078** · September 27th 2012, 08:08 AM

Given y' + 3y = 5te^-t. How do I get the integration factor since there is no (t) value for P(t) where (dy/dt) + P(t) = F(t). Do i divided through by t since there is a t on the righthand side? Thanks to all that for your help!

**Soroban** · September 27th 2012, 08:34 AM

Hello, nivek0078!

If you truly understand the Integrating Factor,
. . exactly where is your difficulty?

Given: . $\frac{dy}{dt} + P(y)\!\cdot\!y \:=\:Q(t)$
. . the integrating factor is: . $I \:=\:e^{\int\!P(y)\,dy}$

$\text{Solve: }\:\tfrac{dy}{dt}+ 3y \:=\: 5te^{-t}$

We have: . $I \;=\;e^{\int\!3\,dt} \;=\;e^{3t}$

Then: . . $e^{3t}\tfrac{dy}{dt} + 3e^{3t}y \;=\;5\:\!t\:\!e^{2t}$

. . . . . . . . . $\frac{d}{dt}\left(e^{3t}y\right) \;=\;5\:\!t\:\!e^{2t}$

. . . . . . . . . . . . $e^{3t}y \;=\;5\!\!\int\!\! t\:\!e^{2t}\,dt$

Can you finish it now?

**nivek0078** · September 27th 2012, 08:44 AM

Yes! Thank you! That was what i was wondering about.

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