Author Topic: Final review question  (Read 395 times)

Meiyi Lu

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Final review question
« on: December 06, 2018, 04:36:27 PM »
Find the general solution of the system of ODEs

$
\left\{
   \begin{array}{c}
      x'_t = -\frac{5}{4}x + \frac{3}{4}y + \frac{2}{1+ e^t},\\
      y'_t = \frac{3}{4}x - \frac{5}{4}y.
   \end{array}
\right.
$

shaohan shi

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Re: Final review question
« Reply #1 on: December 06, 2018, 04:43:42 PM »
$\left( \begin{array}{c}
x'\\
y'
\end{array}
 \right) =
 \left( \begin{array}{cc}
 -\frac{5}{4} & \frac{3}{4}\\
 \frac{3}{4} &  -\frac{5}{4}
  \end{array} \right)\left( \begin{array}{c}
x\\
y
\end{array}
 \right) + \left( \begin{array}{c}
\frac{2}{1+ e^t}\\
0
\end{array}
 \right)
$

$
x'  = Ax + g(t)
$

\textcircled{1} homo: $x' = Ax$ det $(A - \lambda I) = 0$ $\lambda_1 = -\frac{1}{2}$ $\lambda_2 = -2$

$(A - \lambda I)x = 0$

$\therefore$ $c_1\left( \begin{array}{c} 1\\1\end{array} \right)e^{-\frac{1}{2}t} + c_2\left( \begin{array}{c} 1\\-1\end{array} \right)e^{2t}$

\quad $e_1 = \left( \begin{array}{c} 1\\1 \end{array} \right)$, $e_2 = \left( \begin{array}{c} 1\\-1 \end{array} \right)$

\textcircled{2} non-homo: $\varphi =\left( \begin{array}{cc} e^{-\frac{1}{2}t} & e^{-2t}\\ e^{-\frac{1}{2}t} & -e^{-2t}\end{array} \right)$

$\therefore$ $\varphi u' = g$

$\therefore$ $\left( \begin{array}{cc} e^{-\frac{1}{2}t} & e^{-2t}\\ e^{-\frac{1}{2}t} & -e^{-2t}\end{array} \right)\left( \begin{array}{c} u'_1 \\ u'_2\end{array} \right) = \left( \begin{array}{c} \frac{2}{1+e^t} \\0 \end{array} \right) $

\quad
$
\left(
\begin{array}{c:c}
 \begin{array}{cc} e^{-\frac{1}{2}t} & e^{-2t}\\ e^{-\frac{1}{2}t} & -e^{-2t}\end{array}&
\begin{array}{c} \frac{2}{1+e^t} \\0 \end{array}
\end{array}
\right)\underline{R_1 x(-1) + R_2}
\left(
\begin{array}{c:c}
 \begin{array}{cc} e^{-\frac{1}{2}t} & e^{-2t}\\ 0 & -2e^{-2t}\end{array}&
\begin{array}{c} \frac{2}{1+e^t} \\\frac{-2}{1+e^t} \end{array}
\end{array}
\right)
$

$\therefore$ $-2e^{-2t} u'_2 = \frac{-2}{1+e^t}$

$\therefore$ $u'_2 = \frac{e^{2t}}{1+e^t}$

$\therefore$ $e^{-\frac{1}{2}t} u'_1 + e^{-2t} u'_2 = \frac{2}{1+e^t}$

$\therefore$  $u'_1 = \frac{e^{\frac{t}{2}}}{1+ e^{t}}$

$\therefore$  $\begin{aligned}u_1 = \int u'_1\ud t = \int \frac{e^{\frac{t}{2}}}