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### Topics - Ranran Wang

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##### Quiz-5 / LEC5101 Quiz 5
« 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\\
\end{eqnarray}
Therefore,\\
$$Y(t)=(t+\dfrac{1}{2})e^{-2t}\cdot e^{t}+(-2e^{-t})\cdot t=(\dfrac{1}{2}-t)e^{-t}$$
Hence, the general solution:\\
\begin{eqnarray}
&&y(t)=y_{c}(t)+Y(t)\\
\end{eqnarray}
Therefore, the particular solution of the given nonhomogeneous equation is\\
$$Y(t)=(\dfrac{1}{2}-t)e^{-t}$$
\end{document}

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##### Quiz-4 / TUT0501 Quiz4
« on: October 18, 2019, 09:44:52 PM »
This is my solution for the question in quiz.

3
##### Quiz-3 / TUT 0501 Quiz 3
« on: October 11, 2019, 03:33:38 PM »
Here is my question and solution for the question I met in Quiz 3.

4
##### Quiz-2 / TUT0501 Quiz2
« on: October 04, 2019, 03:22:06 PM »
This is the question I do in my quiz 2 and my solution for the question.

5
##### Quiz-1 / TUT 0501 Quiz 1
« on: September 27, 2019, 06:41:29 PM »
These are my answers for my quiz.

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