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Quiz-5 / Quiz5 Victor's section Lec5101
« on: November 01, 2019, 02:00:01 PM »
$$
y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t} \quad t>0
$$
$
\begin{array}{c}{r^{2}+4 r+4=0} \\ {(r+2)^{2}=0} \\ {r=-2,-2}\end{array}
$
Since the roots of characteristic equation are real and repeating.
Therefore, the complimentary solution of differential equation is as follows:
$y_{e}(t)=c e^{-2 t}+c_{2} t e^{-2 t}$
From complimentary solution, the two fundamental solutions of the differential equation are
$y_{1}(t)=e^{-2 t}$ and $y_{2}(t)=t e^{-2 t}$.
Apply the method of variation of parameters to determine the particular solution $Y(t)$
Determine the Wronskian as follows:
Determine the Wronskian as follows:
$
\begin{aligned} W\left(y_{1}, y_{2}\right)(t) &=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=e^{-2 t}\left(-2 t e^{-2 t}+e^{-2 t}\right)-\left(-2 e^{-2 t}\right)\left(t e^{-2 t}\right) \\ &=-2 t e^{-t^{4}}+e^{-4 t}+t e^{-4 t} \end{aligned}
$
$W\left(y_{1}, y_{2}\right)(t)=e^{-4 t}$
By the method of variation of parameters, the particular solution is given as follows:
$Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$
Here, $u_{1}(t)$ and $u_{2}(t)$ are the parameters defined as follows:
$u_{1}(t)=-\int \frac{y_{2}\left(t^{2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$ and $u_{2}(t)=\int \frac{y_{1}\left(t^{2} e^{-t^{2}}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$
$\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}\left(r^{-2} e^{-z}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=-\int \frac{t e^{-2 t}-f^{2} e^{-z t}}{e^{-4}} d t \\ &=-\int \frac{r^{-2} e^{-t}}{e^{-t}} d t \\ &=-\int t^{-1} d t \\=&-\ln t \end{aligned}$ $\begin{aligned} u_{2}(t) &=\int \frac{y_{1}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=\int \frac{e^{-2 t} t^{-2} e^{-t}}{e^{-4 t}} d t \\ &=\int \frac{e^{-t^{2} t^{-2}}}{e^{-4 t}} d t \\ &=\int t^{-2} d t \\ &=-t^{-1} \end{aligned}$
Substitute all values in equation $Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$ for particular solution.
$
\begin{aligned} Y(t) &=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t) \\ &=(-\ln t)\left(e^{-2 t}\right)+\left(t e^{-2 t}\right)\left(-t^{-1}\right) \\ &=-e^{-2 t} \ln t-e^{-2 t} \end{aligned}
$
Add the complimentary solution and particular solution to determine the general solution of the
differential equation as follows:
$\begin{aligned} y &=y_{e}(t)+Y(t) \\ &=c e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t} \\ &=(c-1) e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \end{aligned}$
y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t} \quad t>0
$$
$
\begin{array}{c}{r^{2}+4 r+4=0} \\ {(r+2)^{2}=0} \\ {r=-2,-2}\end{array}
$
Since the roots of characteristic equation are real and repeating.
Therefore, the complimentary solution of differential equation is as follows:
$y_{e}(t)=c e^{-2 t}+c_{2} t e^{-2 t}$
From complimentary solution, the two fundamental solutions of the differential equation are
$y_{1}(t)=e^{-2 t}$ and $y_{2}(t)=t e^{-2 t}$.
Apply the method of variation of parameters to determine the particular solution $Y(t)$
Determine the Wronskian as follows:
Determine the Wronskian as follows:
$
\begin{aligned} W\left(y_{1}, y_{2}\right)(t) &=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=e^{-2 t}\left(-2 t e^{-2 t}+e^{-2 t}\right)-\left(-2 e^{-2 t}\right)\left(t e^{-2 t}\right) \\ &=-2 t e^{-t^{4}}+e^{-4 t}+t e^{-4 t} \end{aligned}
$
$W\left(y_{1}, y_{2}\right)(t)=e^{-4 t}$
By the method of variation of parameters, the particular solution is given as follows:
$Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$
Here, $u_{1}(t)$ and $u_{2}(t)$ are the parameters defined as follows:
$u_{1}(t)=-\int \frac{y_{2}\left(t^{2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$ and $u_{2}(t)=\int \frac{y_{1}\left(t^{2} e^{-t^{2}}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$
$\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}\left(r^{-2} e^{-z}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=-\int \frac{t e^{-2 t}-f^{2} e^{-z t}}{e^{-4}} d t \\ &=-\int \frac{r^{-2} e^{-t}}{e^{-t}} d t \\ &=-\int t^{-1} d t \\=&-\ln t \end{aligned}$ $\begin{aligned} u_{2}(t) &=\int \frac{y_{1}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=\int \frac{e^{-2 t} t^{-2} e^{-t}}{e^{-4 t}} d t \\ &=\int \frac{e^{-t^{2} t^{-2}}}{e^{-4 t}} d t \\ &=\int t^{-2} d t \\ &=-t^{-1} \end{aligned}$
Substitute all values in equation $Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$ for particular solution.
$
\begin{aligned} Y(t) &=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t) \\ &=(-\ln t)\left(e^{-2 t}\right)+\left(t e^{-2 t}\right)\left(-t^{-1}\right) \\ &=-e^{-2 t} \ln t-e^{-2 t} \end{aligned}
$
Add the complimentary solution and particular solution to determine the general solution of the
differential equation as follows:
$\begin{aligned} y &=y_{e}(t)+Y(t) \\ &=c e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t} \\ &=(c-1) e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \end{aligned}$