# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:29:43 AM

Title: TT1 Problem 4 (morning)
Post by: Victor Ivrii on October 16, 2018, 05:29:43 AM
Find the general solution for equation
\begin{equation*}
y''(t)+4y'(t)+5y(t)=e^{-2t} +8\sin(t) .
\end{equation*}
Title: Re: TT1 Problem 4 (morning)
Post by: Jingze Wang on October 16, 2018, 08:29:02 AM
At first, find complement solution
Let y''(t)+4y'(t)+5y(t)=0
we get r^2+4r+5=0
r=-2+i and -2-i
Yc=e^(-2t)(C1 cost+C2 sint)
Then, find particular solution
Let y''(t)+4y'(t)+5y(t)=e^(-2t)
Let Yp1=Ae^(-2t)
Yp1'=-2Ae^(-2t)
Yp1''=4Ae^(-2t)
Plug in we get A=1
Let y''(t)+4y'(t)+5y(t)=8sint
Let Yp2=Bcost+Csint
Yp2'=-Bsint+Ccost
Yp2''=-Bcost-Csint
Plug in, we get B=-1, C=1
Finally, the general solution is Y=e^(-2t)(C1 cost+C2 sint)+e^(-2t)-cost+sint
Title: Re: TT1 Problem 4 (morning)
Post by: Qi Cui on October 16, 2018, 05:06:00 PM
The solution attached below.
Title: Re: TT1 Problem 4 (morning)
Post by: Samarth Agarwal on October 17, 2018, 02:35:44 PM
First we make the equation into the characteristic equation to find the roots:
$$y''(t) + 4y'(t) + 5y(t) = e^{-2t} + 8sin(t)$$
$$\mbox{Therefore, } r^{2} + 4r + 5 = 0$$
$$\mbox{Therefore, } r = -2 \pm i$$
$$\mbox{Therefore, } y(t) = c_1e^{-2t}cos(t)+ c_2e^{-2t}sin(t)$$
Now we change to non homogeneous equation and try to find the value of A:
$$y''(t) + 4y'(t) + 5y(t) = e^{-2t}$$
Let   $$y(t) = Ae^{-2t}$$
$$\mbox{Therefore, } y'(t) = -2Ae^{-2t} \mbox{ and } y''(t) = 4Ae^{-2t}$$
$$\mbox{Therefore, } 4Ae^{-2t} + 4(-2Ae^{-2t}) + 5Ae^{-2t} = e^{-2t}$$
$$Ae^{-2t} = e^{-2t}$$
$$A = 1$$
Now change to non homogenous equation and find the value of B and C:
$$y''(t) + 4y'(t) + 5y(t) = 8sin(t)$$
Let   $$y(t) = Bcos(t) + Csin(t)$$
$$\mbox{Therefore, } y'(t) = -Bsin(t) + Ccos(t) \mbox{ and } y''(t) = -Bcos(t) -Csin(t)$$
$$\mbox{Therefore, } -Bcos(t) -Csin(t) + 4(-Bsin(t) + Ccos(t)) + 5(Bcos(t) + Csin(t)) = 8sin(t)$$
$$\mbox{Therefore, } -Bcos(t) -Csin(t) + -4Bsin(t) + 4Ccos(t) + 5Bcos(t) + 5Csin(t) = 8sin(t)$$
$$\mbox{Therefore, } 4Bcos(t) + 4Csin(t) - 4Bsin(t) + 4Ccos(t) = 8sin(t)$$
$$sin(t)(4C - 4B) + cos(t)(4B + 4C) = 8sin(t)$$
$$\mbox{Therefore, } 4C - 4B = 8 \mbox{ and } 4B + 4C = 0$$
$$\mbox{Therefore, } B = -1 \mbox{ and } C = 1$$
$$\mbox{Therefore, } y(t) = c_1e^{-2t}cos(t)+c_2e^{-2t}sin(t) + e^{-2t} + sin(t) - cos(t)$$
Title: Re: TT1 Problem 4 (morning)
Post by: Victor Ivrii on October 18, 2018, 04:31:11 AM
Jingze  solved right but the typing was atrocious!

Samarth did everything right, but sin, cos etc must be escaped: \sin x producing $\sin x$

Qi, no reason for you to post, except the bad typing of Jingze