Author Topic: TUT0302  (Read 627 times)

annielam

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TUT0302
« on: September 27, 2019, 02:02:07 PM »
$y'-y=2te^{2t}$

$p(t)=-1$
$g(t)=2te^{2t}$
$p(t)= e^{\int -1 dt} = e^{-t}$

Multiply Both Side

$e^{-t}y'-e^{-t}y = 2te^{2t}e^{-t}$
$\frac{d}{dt}$$[e^{-t}y] = 2te^{t}$
$e^{-t}y=\int 2te^tdt $

By Part

$u=2t$
$dv=e^tdt$
$du=2dt$
$v=e^t$
${\int 2te^tdt}=2te^t-{\int 2e^tdt}=2te^t-2e^t+C=2e^t(t-1)+C $

$e^{-t}y=2e^t(t-1)+C$

$y=2e^{2t}(t-1)+Ce^t$

$1=y(0)=2e^0(0-1)+Ce^0=-2+C$
$C=3$

General Solution:
$y=2e^{2t}(t-1)+3e^t$
« Last Edit: September 27, 2019, 02:29:09 PM by annielam »