Author Topic: TUT0302  (Read 451 times)

annielam

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TUT0302
« on: October 18, 2019, 02:12:38 PM »
Find General Solution:
$y''+2y'+y=2e^{-t}$


$r^2+2r+r=0$
$(r+1)(r+1)=0$
$r=-1,-1$

$y=c_{1}e^{-t}+c_2te^{-t}$
$y_{p}(t)=Ae^{-t} \Rightarrow Ate^{-t} \Rightarrow At^2e^{-t}$
$            =$

$y_{p}'(t)=A2te^{-t}+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}+2Ate^{-t}+2At-e^{-t}+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}-e^t2At-e^{-t}2At+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}-4Ae^{-t}+At^2e^{-t}$

$Y=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+2(A2te^{-t}+At^2(-1)(e^{-t}))+At^2e^{-t}$
$Y=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+4Ate^{-t}-2At^2e^{-t}+At^2e^{-t}=2e^{-t}$

$=2Ae^{-t}=2e^{-t}$
$A=1$

$Y=C_1e^{-t}+C_2te^{-t}+t^2e^{-t}$