# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Ranran Wang on November 01, 2019, 02:09:40 PM

Title: LEC5101 Quiz 5
Post by: Ranran Wang on November 01, 2019, 02:09:40 PM
5. Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.\\

\begin{eqnarray*}
&& (1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1 \\
&& y_{1}(t)=e^{t}, y_{2}(t)=t.
\end{eqnarray*}

$$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1;y_{1}(t)=e^{t}, y_{2}(t)=t$$
Hence,

$\left\{ \begin{array}{l} y_{1}(t)=e^{t} \\ y_{1}'(t)=e^{t} \\ y_{1}''(t)=e^{t} \end{array} \right.$
and
$\left\{ \begin{array}{l} y_{2}(t)=t\\ y_{2}'(t)=1\\ y_{2}''(t)=0 \end{array} \right.$

Substitute back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
\ \\
Verified that $y_{1}(t)$ and $y_{2}(t)$ both satisfy the corresponding homogeneous equation.\\
And the complementary solution $y_{c}(t)=c_{1}e^{t}+c_{2}$\\
Now divide both sides of the original equation by $1-t$:\\
$$y''+\dfrac{t}{1-t}-\dfrac{1}{1-t}=-2(t-1)e^{-t}$$
\ \\
Then\\
$$p(t)=\dfrac{t}{1-t},q(t)=-\dfrac{1}{1-t},g(t)=-2(t-1)e^{-t}$$
$$W[y_{1},y_{2}](t)= \left| \begin{array}{cc} y_{1}(t) & y_{2}(t) \\ y_{1}'(t) & y_{2}'(t) \end{array} \right| =(1-t)e^{t}$$
Since the particular solution has the form:\\
$$Y(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t)$$
and\\

\begin{eqnarray}
&&u_{1}(t)=-\int \dfrac{y_{2}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
\end{eqnarray}
\begin{eqnarray}
&&u_{2}(t)=\int \dfrac{y_{1}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
$$Y(t)=(t+\dfrac{1}{2})e^{-2t}\cdot e^{t}+(-2e^{-t})\cdot t=(\dfrac{1}{2}-t)e^{-t}$$
$$Y(t)=(\dfrac{1}{2}-t)e^{-t}$$