# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:26:08 AM

Title: TT1 Problem 4 (main)
Post by: Victor Ivrii on October 16, 2018, 05:26:08 AM
Find the general solution for equation
\begin{equation*}
y''(t)+2y'(t)+5y(t)=8e^{-t} +34\sin(2t) .
\end{equation*}
Title: Re: TT1 Problem 4 (main)
Post by: Christopher Mohanlall on October 16, 2018, 08:46:34 AM
First find the complementary solution for the homogenous equation:
y''(t)+2y'(t)+5y(t) = 8e-t +34sin(2t)

We use the characteristic equation:
r2+2r+5=0

Finding the roots we get:
-1+/-2i

Therefore our yc= c1e-tcos(2t) + c2e-tsin(2t)

Particular Solutions:
8e-t:

Assume: y=Ae-t then:
y'=-Ae-t
y'' =Ae-t

Then:
Ae-t-2Ae-t+5Ae-t=8e-t
A-2A+5A=8
4A=8
A=2

34sin(2t):
Assume y=K1cos(2t)+K2sin(2t)
y'=-2K1sin(2t)+2K2cos(2t)
y''=-4K1cos(2t)-4K2sin(2t)

Then:
-4K1cos(2t)-4K2sin(2t)-4K1sin(2t)+4K2cos(2t)+5K1cos(2t)+5K2sin(2t)=34sin(2t)

Break into components:
Cosine:

-4K1+4K2+5K1=0
4K2+K1=0

K1=-4K2

Sine:

-4K2-4K1+5K2=34
K2-4K1=34

K2-4(-4K2)=34
17K2=34
K2=34/17
K2=2

now:
K1=-4(2)
K1=-8

Therefore:
y= c1e-tcos(2t) + c2e-tsin(2t) + 2e-t + -8cos(2t)+2sin(2t)
Title: Re: TT1 Problem 4 (main)
Post by: Jerry Qinghui Yu on October 16, 2018, 09:14:51 AM
Let me retype it in MathJax
First solve homogenous complement equation
\begin{align*}
y'' + 2y' + 5y &= 0\\
r^2 + 2r + 5 &= 0\\
r_1 = 1+2i,\ &r_2 = 1-2i
\end{align*}
And
\begin{align}
y_c = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t)
\end{align}
Now solve non homogenous part
\begin{align*}
y'' + 2y' + 5y &= 8e^{-t}\\
\end{align*}
Let $Y_1 = Ae^{-t}$, then $Y_1' = -Ae^{-t},\ Y_1'' = Ae^{-t}$
\begin{align*}
A - 2A + 5A &= 8\\
A &= 2\\
Y_1 = 2e^{-t}
\end{align*}
And
\begin{align*}
y'' + 2y' + 5y &= 34\sin(2t)\\
\end{align*}
Let
\begin{align*}
Y_2 &= B\cos(2t) + C\sin(2t)\\
Y_2' &= 2C\cos(2t) - 2B\sin(2t)\\
Y_2'' &= -4B\cos(2t) - 4C\sin(2t)\\
\end{align*}
and
\begin{align*}
-4B\cos(2t) - 4C\sin(2t) +  4C\cos(2t) - 4B\sin(2t) + 5B\cos(2t) + 5C\sin(2t) &= 34\sin(2t)\\
B\cos(2t) + C\sin(2t)  + 4C\cos(2t) - 4B\sin(2t) &= 34\sin(2t)
\end{align*}
Solve the linear equation
\begin{align*}
B+4C&=0\\
C-4B&=34\\
B = -8 &,\ C=2\\
Y_2 = 2\sin(2t)&-8\cos(2t)
\end{align*}
Combining all we have
\begin{align*}
y = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t) + 2e^{-t} + 2\sin(2t)-8\cos(2t)
\end{align*}
Title: Re: TT1 Problem 4 (main)
Post by: Victor Ivrii on October 18, 2018, 04:18:23 AM
Christopher  solved right but typing was both painful and funny looking (don't use html forum tags, use LaTeX/MathJax).

Jerry did everything right, but was not the first.