next up previous
Next: About this document ...

MAT 275 (Review)

TEST 2

Instr. S.Nikitin

Name_______________

ASU-ID#_______________

Solve the following initial value problems.

1.

\begin{displaymath}
(\frac{d}{dt})^4 y + 2 (\frac{d}{dt})^3 y + 2 (\frac{d}{dt})^2 y = 3e^t + 2te^{-t} + e^{-t}\sin(t)
\end{displaymath}


\begin{displaymath}
y(0)=\frac{d}{dt}y(0) = (\frac{d}{dt})^2y(0)= (\frac{d}{dt})^3 y(0) = 0
\end{displaymath}

(Use the method of undetermined coefficients).







2.

\begin{displaymath}
(\frac{d}{dt})^3 y - (\frac{d}{dt})^2 y + (\frac{d}{dt}) y - y = f(t) ,
\end{displaymath}

where

\begin{displaymath}
f(t)=\left\{\begin{array}{ccc}
0 & \mbox{ for } & 0\le t ...
... t <2\\
0 & \mbox{ for } & t \ge 2
\end{array}
\right.
\end{displaymath}


\begin{displaymath}
y(0)=2,\;\;\;\frac{d}{dt}y(0) = -1,\;\;\;\; (\frac{d}{dt})^2y(0)= 1
\end{displaymath}

(Use Duhamel' integral)







3.

\begin{displaymath}
(\frac{d}{dt})^2 y + 5(\frac{d}{dt}) y + 6y = te^{-2t} + \cos(t)
\end{displaymath}


\begin{displaymath}
y(0)=1,\;\;\;\frac{d}{dt}y(0) = 0
\end{displaymath}

(Use the method of undetermined coefficients).







4.

\begin{displaymath}
(\frac{d}{dt})^2 y - 2(\frac{d}{dt}) y + 2y = f(t) ,
\end{displaymath}

where

\begin{displaymath}
f(t)=\left\{\begin{array}{ccc}
0 & \mbox{ for } & 0\le t ...
...t <3 \\
0 & \mbox{ for } & t\ge 3
\end{array}
\right.
\end{displaymath}


\begin{displaymath}
y(0)=1,\;\;\;\frac{d}{dt}y(0) = 1
\end{displaymath}

(Use Duhamel' integral)







5. Find the general solution of the linear system


\begin{displaymath}
\frac{dx}{dt}=
\left(
\begin{array}{cc}
2 & 2 \\
6 & -2
\end{array}
\right)
x
\end{displaymath}





Sergey Nikitin 2007-10-12