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yangqi40

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Quiz 5 Lecture5101
« on: November 01, 2019, 02:16:23 AM »
Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation then find a particular solution of the given nonhomogeneous equation


$(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1$

$y_{1}(t)=e^{t}, y_{2}(t)=t$

$(1-t) y^{\prime \prime}+t y-y=2(t-1)^{2} e^{-t}, 0<t<1 ; y_{1}(t)=e^{t}, y_{2}(t)=t$




$\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t}} \\ {y_{1}^{\prime \prime}(t)=e^{t}}\end{array}\right.$ and $\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.$


$\qquad$Substitute back into the homogeneous equation


$(1-t) y^{\prime \prime}+t y^{\prime}-y=0$

$y^{\prime \prime}+\frac{t}{1-t}-\frac{1}{1-t}=-2(t-1) e^{-t}$
$\qquad$Then


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



$W\left[y_{1}, y_{2}\right](t)=\left|\begin{array}{ll}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|=(1-t) e^{t}$


Since the particular solution has the form:


$Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$


and

\begin{equation}
\label{eq:D1}
    u_{1}(t)=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t
\end{equation}

\begin{equation}
\label{eq:D2}
    =-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t
\end{equation}

\begin{equation}
\label{eq:D3}
    =-2 \int t e^{-2 t} d t
\end{equation}

\begin{equation}
\label{eq:D4}
   =\left(t+\frac{1}{2}\right) e^{-2 t}
\end{equation}

\begin{equation}
\label{eq:D5}
   u_{2}(t)=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t
\end{equation}

\begin{equation}
\label{eq:D6}
   =\int \frac{e^{t} \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t
\end{equation}

\begin{equation}
\label{eq:D7}
   =2 \int e^{-t}
\end{equation}

\begin{equation}
\label{eq:D8}
   =-2 e^{-t}
\end{equation}

Therefore,


$Y(t)=\left(t+\frac{1}{2}\right) e^{-2 t} \cdot e^{t}+\left(-2 e^{-t}\right) \cdot t=\left(\frac{1}{2}-t\right) e^{-t}$


The general solution is:

\begin{equation}
\label{eq:D9}
   y(t)=y_{c}(t)+Y(t)
\end{equation}

\begin{equation}
\label{eq:D10}
   =c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t}
\end{equation}

Therefore,the particular solution of the given nonhomogeneous equation is


$Y(t)=\left(\frac{1}{2}-t\right) e^{-t}$