 # y_1'=-3y_1+y_2+u(t-1)e^t, y_2'=-4y_1+2y_2+u(t−1)e^t, y_1(0)=0, y_2(0)=21 vazelinahS 2020-11-01 Answered

${y}_{1}^{\prime }=-3{y}_{1}+{y}_{2}+u\left(t-1\right){e}^{t},{y}_{2}^{\prime }=-4{y}_{1}+2{y}_{2}+u\left(t-1\right){e}^{t},{y}_{1}\left(0\right)=0,{y}_{2}\left(0\right)=21$
Enclose arguments of functions, numerators, and denominators in parentheses.

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Given: ${y}_{1}^{\prime }=-3{y}_{1}+{y}_{2}+u\left(t-1\right){e}^{t},{y}_{2}^{\prime }=-4{y}_{1}+2{y}_{2}+u\left(t-1\right){e}^{t},{y}_{1}\left(0\right)=0,{y}_{2}\left(0\right)=21$
Apply Laplace transform:
$s{Y}_{1}\left(s\right)-{y}_{1}\left(0\right)=-3{Y}_{1}\left(s\right)+{Y}_{2}\left(s\right)+\frac{{e}^{-\left(s-1\right)}}{s-1}$
$\left(s+3\right){Y}_{1}\left(s\right)-{Y}_{2}\left(s\right)=\frac{{e}^{-\left(s-1\right)}}{s-1}$
$s{Y}_{2}\left(s\right)-{y}_{2}\left(0\right)=-4{Y}_{1}\left(s\right)+2{Y}_{2}\left(s\right)+\frac{{e}^{-\left(s-1\right)}}{s-1}$
$s{Y}_{2}\left(s\right)-21=-4{Y}_{1}\left(s\right)+2{Y}_{2}\left(s\right)+\frac{{e}^{-\left(s-1\right)}}{s-1}$
$\left(s+4\right){Y}_{1}\left(s\right)-2{Y}_{2}\left(s\right)=\frac{{e}^{-\left(s-1\right)}}{s-1}+21$
Solve the equations:
$\left(s+3\right){Y}_{1}\left(s\right)-{Y}_{2}\left(s\right)=\frac{{e}^{-\left(s-1\right)}}{s-1}\cdot \left(-2\right)$
$\left(s+4\right){Y}_{1}\left(s\right)-2{Y}_{2}\left(s\right)=\frac{{e}^{-\left(s-1\right)}}{s-1}+21$
$\left[-2\left(s+3\right)+\left(s+4\right)\right]{Y}_{1}\left(s\right)=-\frac{{e}^{-\left(s-1\right)}}{s-1}+21$
${Y}_{1}\left(s\right)=\frac{{e}^{-\left(s-1\right)}}{\left(s-1\right)\left(s+2\right)}-\frac{21}{\left(s+2\right)}$

Apply inverse Laplace transform:
${y}_{1}\left(t\right)=\left[u\left(t-1\right)\right]\ast {e}^{t}-\left[u\left(t+2\right)\right]\ast {e}^{t+3}-21\ast {e}^{-2t}$
Solution:
$\left(s+3\right)\left(\frac{{e}^{-\left(s-1\right)}}{s-1}-{e}^{3}\left(\frac{{e}^{-\left(s+2\right)}}{s+2}\right)-\frac{21}{s+2}\right)-{Y}_{2}\left(s\right)=\frac{{e}^{-\left(s-1\right)}}{s-1}$
${Y}_{2}\left(s\right)=\left(s+3\right)\frac{{e}^{-\left(s-1\right)}}{\left(s-1\right)}-{e}^{3}\left(s+3\right)\left(\frac{{e}^{-\left(s+2\right)}}{s+2}\right)-\left(s+3\right)\frac{21}{\left(s+2\right)}-\frac{{e}^{-\left(s-1\right)}}{s-1}$