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MAT 274 (Review)

TEST 3

Instr. S.Nikitin

Name_______________

ASU-ID#_______________

I. Solve the following initial value problems.

1.

\begin{displaymath}
\frac{d}{dt}y = \left( \begin{array}{cc}
-8 & 18 \\
-3...
...\left( \begin{array}{c}
e^{-t}\\
1
\end{array} \right)
\end{displaymath}


\begin{displaymath}
y(0)=\left( \begin{array}{c}
4\\
1
\end{array} \right)
\end{displaymath}














2.

\begin{displaymath}
\frac{d}{dt}y = \left( \begin{array}{cc}
-2 & 1 \\
1 &...
...eft( \begin{array}{c}
e^t\\
e^{-t}
\end{array} \right)
\end{displaymath}


\begin{displaymath}
y(0)=\left( \begin{array}{c}
4\\
1
\end{array} \right)
\end{displaymath}














II. Find the general solution, sketch the phase portrait.

a.
$
\frac{d}{dt}y = \left( \begin{array}{cc}
-3 & 1 \\
-5 & 1
\end{array} \right)
$
b.
$
\frac{d}{dt}y = \left( \begin{array}{cc}
-1 & 1 \\
-5 & 1
\end{array} \right)
$
c.
$
\frac{d}{dt}y = \left( \begin{array}{cc}
-2 & \frac13 \\
-3 & -2
\end{array} \right)
$

















III. Find the solution in the integral form (you do not need to calculate the integrals).


\begin{displaymath}
\frac{d}{dt}y = \left( \begin{array}{cc}
0 & 1 \\
-25 ...
...in{array}{c}
\sqrt{sin^2(t)} \\
t
\end{array} \right)
\end{displaymath}


\begin{displaymath}
y(0)=\left( \begin{array}{c}
1\\
1
\end{array} \right)
\end{displaymath}














IV. Find the solution with the help of Laplace transforms.

\begin{displaymath}
\ddot y(t) + 3 \dot y + 2 y = \sin(t)
\end{displaymath}


\begin{displaymath}
y(0)=-1,\;\;\dot y(0)=1
\end{displaymath}




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Sergey Nikitin 2007-04-16