Author Topic: LEC5101 Quiz5  (Read 490 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,

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