# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Lan Cheng on November 01, 2019, 02:00:17 PM

Title: LEC 5101 QUIZ 5
Post by: Lan Cheng on November 01, 2019, 02:00:17 PM
Find the general solution of the given differential equation:

$y"+4y'+4y=t^{-2}e^{-2t},t>0.$

Let $y"+4y'+4y=0.$

$r^{2}+4r+4=0,r_{1}=-2=r_{2}.$

Thus,

$y_{c}(t)=C_{1}e^{-2t}+C_{2}te^{-2t}.$

$W=\begin{vmatrix}e^{-2t} & te^{-2t}\\ -2e^{-2t} & -2te^{-2t}+e^{-2t} \end{vmatrix}=-2te^{4t}+e^{4t}+2te^{4t}=e^{4t}.$

$W_{1}=\begin{vmatrix}0 & te^{-2t}\\ 1 & -2te^{-2t}+e^{-2t} \end{vmatrix}=-te^{-2t}.$

$W_{2}=\begin{vmatrix}e^{-2t} & 0\\ -2e^{-2t} & 1 \end{vmatrix}=e^{-2t}.$

Let $\mu_{1}=\int\frac{W_{1}(t)g(t)}{W(t)}dt=\int\frac{-te^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-ln(t).$

$\mu_{2}=\int\frac{W_{2}(t)g(t)}{W(t)}dt=\int\frac{e^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-\frac{1}{t}.$

Therefore,

$y(t)=y_{1}(t)\mu_{1}+y_{2}(t)\mu_{2}+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$