Author Topic: TUT0301 LEC 5101 QUIZ 5  (Read 3626 times)

LLY

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TUT0301 LEC 5101 QUIZ 5
« on: November 04, 2019, 05:08:13 PM »
\begin{equation}
\begin{array}{l}{(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}\end{array}
\end{equation}
\begin{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
\end{equation}
\begin{equation}
\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t} \text { and }\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.}\end{array}\right.
\end{equation}
\begin{equation}
(1-t) y^{\prime \prime}+t y^{\prime}-y=0
\end{equation}
\begin{equation}
y^{\prime \prime}+\frac{t}{1-t}-\frac{1}{1-t}=-2(t-1) e^{-t}
\end{equation}
\begin{equation}
p(t)=\frac{t}{1-t}, q(t)=-\frac{1}{1-t}, g(t)=-2(t-1) e^{-t}
\end{equation}
\begin{equation}
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}
\end{equation}
\begin{equation}
Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)
\end{equation}
\begin{equation}
\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=-2 \int t e^{-2 t} d t \\ &=\left(t+\frac{1}{2}\right) e^{-2 t} \end{aligned}
\end{equation}
\begin{equation}
\begin{aligned} u_{2}(t) &=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=\int \frac{e^{t} \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=2 \int e^{-t} \\ &=-2 e^{-t} \end{aligned}
\end{equation}
\begin{equation}
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}
\end{equation}
\begin{equation}
\begin{aligned} y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t} \end{aligned}
\end{equation}