Author Topic: Lec5101  (Read 89 times)

Yuefan Wang

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Lec5101
« on: November 15, 2019, 02:00:01 PM »
Find the general solution
$$
\begin{array}{c}{x^{\prime}=\left(\begin{array}{cc}{4} & {-3} \\ {8} & {-6}\end{array}\right) x} \\ {\left(\begin{array}{cc}{4-\lambda} & {-3} \\ {8} & {-6-x}\end{array}\right)\left(\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right)=\left(\begin{array}{c}{0} \\ {0}\end{array}\right)} \\ {\left(\begin{array}{cc}{4-\lambda} & {-3} \\ {8} & {-6-\lambda}\end{array}\right)=0} \\ {r(r+2)=0} \\ {r_{1}=0, \quad r_{2}=-2}\end{array}
$$

when $r_{1}=0$
$$
\begin{array}{c}{\left(\begin{array}{cc}{4} & {-3} \\ {8} & {-6}\end{array}\right)\left(\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)} \\ {x_{1}=\left(\begin{array}{l}{3} \\ {4}\end{array}\right)}\end{array}
$$
$$
\begin{array}{c}{\left(\begin{array}{cc}{6} & {-3} \\ {8} & {-4}\end{array}\right)\left(\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right)=\left(\begin{array}{c}{0} \\ {0}\end{array}\right)} \\ {\lambda_{2}=\left(\begin{array}{l}{1} \\ {2}\end{array}\right)}\end{array}
$$
$$
x^{\prime}(t)=\left(\begin{array}{l}{3} \\ {4}\end{array}\right) \quad x^{2}(t)=\left(\begin{array}{l}{1} \\ {2}\end{array}\right)t^{-2}
$$

general solution
$$
x=c_{1}\left(\begin{array}{l}{3} \\ {4}\end{array}\right)+c_{2}\left(\begin{array}{l}{1} \\ {2}\end{array}\right) t^{-2}
$$

Wang Jingyao

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Re: Lec5101
« Reply #1 on: December 03, 2019, 06:37:22 PM »
I think the general solution should be $x(t)=$$c_1
\left [
\begin{matrix}
3 \\
4
\end{matrix}
\right ]
+$$c_2 e^{-2t}
\left [
\begin{matrix}
1 \\
2
\end{matrix}
\right ]
$