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### Messages - Yeming Wen

Pages: 1 [2]
16
##### TT1 / Re: TT1-problem 3
« on: October 09, 2014, 09:27:17 AM »
First, we solve
y'' - y' - 6y = 0 \label{A}

Notice that the characteristic equation is
\begin{equation*}
r^{2} - r - 6 = 0.
\end{equation*}
Then the solution to the (\ref{A}) is
\begin{equation*}
y =C_1 e^{3t}+ C_2 e^{-2t}
\end{equation*}
Now we need to find a particular solution to

y'' - y' - 6y =10 e^{-2t}- 6
\label{B}

We proceed by undetermined coefficients. Since $-2$  is a root of (\ref{A}) and $0$ is not , we guess
\begin{equation*}
y=Ate^{-2t}+B
\end{equation*}
Then
\begin{equation*}
y'=Ae^{-2t}-2Ate^{-2t}
\end{equation*}
and
\begin{equation*}
y''=-4Ae^{-2t}+4Ate^{-2t}.
\end{equation*}
Then
\begin{equation*}
y'' - y' - 6y = -Ae^{-2t}-6B
\end{equation*}
Compare with
\begin{equation*} 10 e^{-2t}- 6.
\end{equation*}
We get
\end{equation*}
So a particular solution to (\ref{B}) is
\begin{equation*}
y=-2te^{-2t} + 1.
\end{equation*}
So the general solution is
\begin{equation*}
y = C_1e^{3t}+C_2e^{-2t}+1-2te^{-2t}
\end{equation*}

17
##### Quiz 2 / Re: Quiz 2 3.4 #14
« on: October 01, 2014, 11:31:36 PM »
I guess the graph looks like this.

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