LEC5101 Quiz 5

[Wang Jingyao]

Verify that the given function $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.

$(1-t)y’’+ty’-y=2(t-1)^2e^{-t},\quad 0<t<1$

$y_1(t)=e^t,\quad y_2(t)=t$

Solution:

$\left\{\begin{array}{l}y_1(t)=e^t\\y_1’(t)=e^t\\y_1’’(t)=e^t\end{array}\right.$

$\left\{\begin{array}{l}y_2(t)=t\\y_2’(t)=1\\y_2’’(t)=0\end{array}\right.$

Substitute back into the homogeneous equation:

$(1-t)y’’+ty’-y=0$

Verify that $y_1(t)=e^t$ and $y_2(t)=t$ satisfy the corresponding homogeneous equation

Divide both sides of the original equation by $1-t$:

$y’’+\dfrac{t}{1-t}-\dfrac{1}{1-t}=-2(t-1)e^{-t}$

Therefore,

$p(t)= \dfrac{t}{1-t},\quad q(t)= -\dfrac{1}{1-t},\quad g(t)= -2(t-1)e^{-t}$

$W[y_1(t),y_2(t)]=$$
\left |
\begin{matrix}
y_1(t) & y_2(t) \\
y_1’(t) & y_2’(t)
\end{matrix}
\right |
$$
=(1-t)e^t$

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

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

Since,

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

Therefore,

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

Therefore, the particular solution of the given nonhomogeneous equation is

$Y(t)= (\dfrac{1}{2}-t)e^{-t}$