### Author Topic: LEC5101 Quiz5  (Read 523 times)

#### Wang Jingyao

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##### LEC5101 Quiz5
« on: November 01, 2019, 12:50:56 AM »
Find the general solution of the given differential equation.

$y’’+4y’+4y=t^{-2}e^{-2t},\quad t>0$

Solution:

For homogeneous equation:

$y’’+4y’+4y=0$

Characteristic equation:

$r^2+4r+4=0$ $\Longrightarrow$ $\left\{\begin{array}{l}r_1=-2\\r_2=-2 \end{array}\right.$

Complementary solution:

$y_c(t)=c_1e^{-2t}+c_2te^{-2t}$

For nonhomogeneous equation $y’’+4y’+4y=t^{-2}e^{-2t}$ we have:

$p(t)=0,\quad q(t)=4,\quad g(t)= t^{-2}e^{-2t}$

Then,

$W[y_1(t),y_2(t)]=$$\left | \begin{matrix} y_1(t) & y_2(t) \\ y_1’(t) & y_2’(t) \end{matrix} \right |$$ =$$\left | \begin{matrix} e^{-2t} & te^{-2t} \\ -2e^{-2t} & -2e^{-2t}+ e^{-2t} \end{matrix} \right |$$ =e^{-4t}$

Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. Therefore,

\begin{align} u_1(t)&=-\int\dfrac{y_2(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\ \notag \\ &=-\int\dfrac{ te^{-2t}\cdot t^{-2}e^{-2t}}{ e^{-4t}}dt\\ \notag \\ &=-\int t^{-1}dt\\ \notag \\ &=-\ln t \end{align}

\begin{align} u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\ \notag \\ &=\int\dfrac{e^{-2t}\cdot t^{-2}e^{-2t}}{ e^{-4t}}dt\\ \notag \\ &=\int t^{-2}dt\\ \notag \\ &=-t^{-1}\\ \end{align}

Since,

$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$

Therefore,

\begin{align} Y(t)&=-\ln t\cdot e^{-2t}+te^{-2t}\cdot (-t^{-1})\notag \\ \notag \\ &=-e^{-2t}\ln t-e^{-2t}\notag\\ \end{align}

Thus, the general solution is,

\begin{align} y(t)&=y_c(t)+Y(t) \notag \\ \notag \\ &= c_1e^{-2t}+c_2te^{-2t}-e^{-2t}\ln t-e^{-2t}\notag \\ \notag \\ &=(c_1-1)e^{-2t}+c_2te^{-2t}-e^{-2t}\ln t\notag \\ \notag \\ y(t)&= c_1e^{-2t}+c_2te^{-2t}-e^{-2t}\ln t\notag \\ \end{align}