MAT244--2019F > Quiz-4

TUT0602 QUIZ4

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Kun Zheng:
Hi everyone for quiz 4 we need to get the original equation of
$y''+y'-6y=12e^{-2t}+12e^{3t}$

First, we get the homogeneous form:
$r^2+r-6=0$
$(r+3)(r-2)=0$
$r1=-3, r2=2$

Then we can get:
$y=C_1e^{-2t}+C_2e^{3t}$
$y'=-2C_1e^{-2t}+3C_2e^{3t}$
$y''=4C_1e^{-2t}+9C_2e^{3t}$

so this means:
$-6B-2B+4B=12$
$-6A+3A+9A=12$
Solve to get
$A=2, B=-3$

So we rewrite:
$2e^{3t}-3e^{-2t}$

Plug in to the solution:
$y=2e^{3t}-3e^{-2t}+C_1e^{2t}+C_2e^{-3t}$