MAT244--2019F > Quiz-5

LEC5101 Quiz5

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Joy Zhou:
$y''+4y'+4y=t^{-2}e^{-2t}, t>0$

The homogeneous differential equation is as follows:
$$y^{\prime\prime}+4 y^{\prime}+4 y=0$$

The characteristic equation for homogeneous differential equation is
\begin{aligned} r^{2}+4 r+4 &=0\\ (r+2)^{2} &=0 \\ r &=-2,-2 \end{aligned}

since the roots of characteristic equation are real and repeating.

Therefore, the solution of differential equation is as follows:
$${y_{c}(t)=c e^{-2 t}+c_{2} te^{-2t}}$$

($c$ and $c_{2}$ are constants)

From above solution, the two fundamental solutions of the differential equation are
$y_{1}(t)=e^{-2 t}$ and $y_{2}(t)=t e^{-2 t}$

Then, use the variation of parameters method to determine the particular solution $Y(t) .$

Find 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}^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {te^{-2t}} \\ {-2 e^{-2 t}} & {-2 t e^{-2 t}+e^{-2 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^{-4 t}+e^{-4 t}+t e^{-4 t} \end{aligned}

$W\left(y_{1}, y_{2}\right)(t)=e^{-4t}$

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 $u_{2}(t)=\int \frac{y_{1}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$

\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=-\int \frac{t e^{-2 t}t^{-2}e^{-2t}}{e^{-4 t}} d t \\ &=-\int \frac{t^{-1} e^{-4 t}}{e^{-4 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^{-2 t}}{e^{-4 t}} d t \\ &=\int \frac{e^{-4 t}t^{-2}}{e^{-4 t}} d t \\ &=\int t^{-2} d t \\ &=-t^{-1} \end{aligned}

Put all the 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}

So, the general solution of the equation is:
$$\begin{array}{l}{y=y_{c}(t)+Y(t)} \\ {=c e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t}} \\ {=(c-1) e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t} \\ {=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t}\end{array}$$

$$(c-1=c_1)$$
Therefore, the required general solution is
$$y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t$$