# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-6 => Topic started by: Yuefan Wang on November 15, 2019, 02:00:01 PM

Title: Lec5101
Post by: Yuefan Wang 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
$$x=c_{1}\left(\begin{array}{l}{3} \\ {4}\end{array}\right)+c_{2}\left(\begin{array}{l}{1} \\ {2}\end{array}\right) t^{-2}$$
Title: Re: Lec5101
Post by: Wang Jingyao 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 ]$