# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 05:12:34 PM

Title: P4-Morning
Post 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*}
Title: Re: P4-Morning
Post by: Vivian Ngo on February 15, 2018, 05:28:27 PM
*I will type up the solutions soon*

(https://scontent-yyz1-1.xx.fbcdn.net/v/t34.0-12/28053566_10213726400345645_1400938461_n.png?oh=6d09f74d06187a4510a0d7675798787f&oe=5A87762C)

*Typed solutions to come soon*
Title: Re: P4-Morning
Post by: Vivian Ngo on February 16, 2018, 12:21:04 AM
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:
$y(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t) + e^{-4t} + 2sin(3t) -3cos(3t)$