Solve the system

$x^{\text{'}}\left[\begin{array}{cc}6& -2\\ 2& 10\end{array}\right]x=\left[\begin{array}{c}{e}^{-2t}\\ 0\end{array}\right],x_1(0)=2$ and $x_2(0)=-1$

by using the given information.

$e^{tP}=\left[\begin{array}{ccc}{e}^{8t}(1-2t)& e^{8t}(-2t)& e^{8t}(12t)\end{array}\right]$

$e^{8t}(2t)$

${\displaystyle \underset{0}{\overset{t}{}}}{e}^{sP}f(s)ds=\left[\begin{array}{c}-\frac{2}{6}t{e}^{6t}\frac{8}{36}{e}^{6t}-\frac{8}{36}\end{array}\right]$

$\frac{2}{6}t{e}^{6t}-\frac{2}{36}{e}^{6t}\frac{2}{36}$

$x_1(t)=$

$x_2(t)=$

Trending videos

Search

This Eattery Dre.