### Author Topic: LEC5101 Quiz 5  (Read 927 times)

#### Yuefan Wang

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##### LEC5101 Quiz 5
« on: November 01, 2019, 02:00:00 PM »
Find the general solution of $y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t}, t>0$

$$\begin{array}{l}{\text { homo: } y^{\prime \prime}+4 y^{\prime}+4 y=0} \\ {\qquad \begin{array}{c}{(r+2)^{2}=0} \\ {r_{1}=-2, r_{2}=-2} \\ {\therefore y_{c}(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}}\end{array}}\end{array}$$
\begin{array}{l}{\text { non-homo: }} \\ {\qquad \begin{aligned} w\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right| &=\left|\begin{array}{cc}{e^{-2 t}} & {te^{-2 t}} \\ {-2 e^{-2 t}} & {-2 t^{-2t}+e^{-2 t} }\end{array}\right| \\ &=-2 t e^{-4 t}+e^{-4 t}+t e^{-4 t}=e^{-4t} \end{aligned}}\end{array}
$$\begin{array}{l}{w_{1}\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc}{0} & {e^{-2 t}} \\ {1} & {-2 t e^{-2 t}+e^{-2 t} |}\end{array}\right|=-t e^{-2 t}} \\ {w_{2}\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc}{e^{-2 t}} & {0} \\ {-2 e^{-2 t}} & {1}\end{array}\right|=e^{-2 t}}\end{array}$$
\begin{aligned} Y(t)&=e^{-2 t} \int \frac{\left(-s e^{-2 s}\right) \cdot\left(s^{-2} e^{-2 s}\right)}{e^{-2 s}} d s+t e^{-2 t} \int \frac{\left(e^{-2 s}\right) \cdot\left(s^{-2} e^{-2 s}\right)}{e^{-2 s}} d s\\ &=e^{-2 t} \int -t^{-1} d t+t e^{-2 t} \int t^{-2} d t\\&=-e^{-2 t} \ln t-e^{-2 t}\\ y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t} \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \end{aligned}