MAT244-2018S > Term Test 1

P4-Morning

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Victor Ivrii:
Find the general solution for equation
\begin{equation*}
y''(t)+8y'(t)+25y(t)=9 e^{-4t}+ 104\sin(3t).
\end{equation*}

Vivian Ngo:
*I will type up the solutions soon*



*Typed solutions to come soon*

Vivian Ngo:
Characteristic equation:
$r^2+8r+25=0$
r = $-4 +3i, -4-3i$ (using quadratic equation)

Homogeneous solution:
$y_c(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t)$

Particular solutions:

First Particular:
$Y = Ae^{-4t}$
$Y' = -4Ae^{-4t}$
$Y'' = 16Ae^{-4t}$

$A(16+8(-4)+25)=9$
$9A=9$
$A=1$

Second Particular:

$Y = Asin(3t)+Bcos(3t)$
$Y' = 3Acos(3t)-3Bsin(3t)$
$Y'' = -9Asin(3t)-9Bcos(3t)$

Plug into the given equation:
sines:
$-9A+8(-3B)+25A = 104$
$16A-24B=104$
$2A-3B=13$

cosines:
$-9B+8(3A)+25B = 0$
$16B+24A=0$
$2B+3A=0$

==> $A=2, B=-3$

General solution:
$y(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t) + e^{-4t} + 2sin(3t) -3cos(3t)$

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