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!
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!
Hello, nivek0078!
If you truly understand the Integrating Factor,
. . exactly where is your difficulty?
Given: .$\displaystyle \frac{dy}{dt} + P(y)\!\cdot\!y \:=\:Q(t)$
. . the integrating factor is: .$\displaystyle I \:=\:e^{\int\!P(y)\,dy} $
$\displaystyle \text{Solve: }\:\tfrac{dy}{dt}+ 3y \:=\: 5te^{-t}$
We have: .$\displaystyle I \;=\;e^{\int\!3\,dt} \;=\;e^{3t}$
Then: . .$\displaystyle e^{3t}\tfrac{dy}{dt} + 3e^{3t}y \;=\;5\:\!t\:\!e^{2t}$
. . . . . . . . . $\displaystyle \frac{d}{dt}\left(e^{3t}y\right) \;=\;5\:\!t\:\!e^{2t}$
. . . . . . . . . . . . $\displaystyle e^{3t}y \;=\;5\!\!\int\!\! t\:\!e^{2t}\,dt$
Can you finish it now?