Toronto Math Forum
MAT2442018S => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 05:12:34 PM

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

*I will type up the solutions soon*
(https://scontentyyz11.xx.fbcdn.net/v/t34.012/28053566_10213726400345645_1400938461_n.png?oh=6d09f74d06187a4510a0d7675798787f&oe=5A87762C)
*Typed solutions to come soon*

Characteristic equation:
$r^2+8r+25=0$
r = $4 +3i, 43i$ (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:
$9A+8(3B)+25A = 104$
$16A24B=104$
$2A3B=13$
cosines:
$9B+8(3A)+25B = 0$
$16B+24A=0$
$2B+3A=0$
==> $A=2, B=3$
General solution:
$y(t) = c_1e^{4t}cos(3t) + c_2e^{4t}sin(3t) + e^{4t} + 2sin(3t) 3cos(3t)$