Author Topic: TT1-problem 3  (Read 905 times)

Victor Ivrii

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TT1-problem 3
« on: October 09, 2014, 02:00:15 AM »
Find the general solution for equation
\begin{equation*}
z''(t)-z'(t)-6z(t)=-6+10 e^{-2t} .
\end{equation*}

Yeming Wen

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Re: TT1-problem 3
« Reply #1 on: October 09, 2014, 09:27:17 AM »
First, we solve
\begin{equation} y'' - y' - 6y = 0 \label{A}
\end{equation}
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
\begin{equation}
y'' - y' - 6y =10 e^{-2t}- 6
\label{B}
\end{equation}
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
\begin{equation*} A=-2 ,\qquad  B=1.
 \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*}
« Last Edit: October 09, 2014, 10:02:43 AM by Victor Ivrii »

Victor Ivrii

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Re: TT1-problem 3
« Reply #2 on: October 09, 2014, 10:03:44 AM »
Good job. I made a minor editing. Pay attention to \label - \ref mechanism