91Ó°ÊÓ

The differential equations are given as:

$\begin{array}{l}{x}^{\text{'}}-y^{\text{'}}=\left(sint\right)u(t-\mathrm{Ï€});\text{ â¶Ä‰â¶Ä‰}x\left(0\right)=1,\\ x+y^{\text{'}}=0;\text{ â¶Ä‰â¶Ä‰}y\left(0\right)=1\end{array}$

Given initial value equations are,

$\begin{array}{c}{x}^{\text{'}}-y^{\text{'}}=sintu(t-\mathrm{Ï€}),\text{ â¶Ä‰â¶Ä‰}x\left(0\right)=1\\ x^{\text{'}}-y^{\text{'}}=-(-sint)u(t-\mathrm{Ï€})\\ =-\left(sin\right(t-\mathrm{Ï€}\left)\right)u(t-\mathrm{Ï€})\mathrm{....}\left(1\right)\\ x+y^{\text{'}}=0,\text{ â¶Ä‰â¶Ä‰}y\left(0\right)=\mathrm{1...}\left(2\right)\end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{c}sx\left(s\right)-x\left(0\right)-sy\left(s\right)+y\left(0\right)=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}\\ sx\left(s\right)-1-sy\left(s\right)+1=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}\\ sx\left(s\right)-sy\left(s\right)=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}\mathrm{....}\left(3\right)\end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{c}x\left(s\right)+sy\left(s\right)-y\left(0\right)=0\\ x\left(s\right)+sy\left(s\right)-1=0\\ x\left(s\right)=1-sy\left(s\right)\mathrm{.....}\left(4\right)\end{array}$

Putting equation fourth into third we get

$\begin{array}{c}s[1-sy(s\left)\right]-sy\left(s\right)=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}\\ s-s^{2}y\left(s\right)-sy\left(s\right)=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}\\ s-(s^2+s)y\left(s\right)=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}\\ -(s^2+s)y\left(s\right)=-\frac{e^{-\mathrm{Ï€}s}}{1+s^2}-s\end{array}$

Simplify further as:

$\begin{array}{c}y\left(s\right)=\frac{e^{-\mathrm{Ï€}s}}{(1+s^2)(s^2+s)}+\frac{s}{(s^2+s)}\\ =\frac{e^{-\mathrm{Ï€}s}}{(1+s^2)s(s+1)}+\frac{1}{(s+1)}\end{array}$

Using partial fraction we get

$\begin{array}{c}y\left(s\right)=e^{-\mathrm{Ï€}s}\left[\frac{-s-1}{2(s^2+1)}+\frac{1}{s}-\frac{1}{2(s+1)}\right]+\frac{1}{(s+1)}\\ =e^{-\mathrm{Ï€}s}\left[\frac{-s}{2(s^2+1)}-\frac{1}{2(s^2+1)}+\frac{1}{s}-\frac{1}{2(s+1)}\right]+\frac{1}{(s+1)}\end{array}$

Taking inverse Laplace transform we get

$\begin{array}{c}y\left(t\right)=\left[\frac{-1cos(t-\mathrm{Ï€})}{2}-\frac{1sin(t-\mathrm{Ï€})}{2}+1-\frac{1e^{-(t-\mathrm{Ï€})}}{2}\right]u(t-\mathrm{Ï€})+e^{-t}\\ =[\frac{cost}{2}+\frac{sint}{2}+1-\frac{1e^{-(t-\mathrm{Ï€})}}{2}]u(t-\mathrm{Ï€})+e^{-t}\end{array}$

From equation fourth

$\begin{array}{c}x\left(s\right)=1-sy\left(s\right)\\ x\left(s\right)=1-s\left[\frac{c^{\mathrm{Ï€}s}}{(1+s^2)s(s+1)}+\frac{1}{(s+1)}\right]\\ =1-\frac{e^{-\mathrm{Ï€}s}}{(1+s^2)(s+1)}+\frac{s}{(s+1)}\\ =-\frac{e^{-\mathrm{Ï€}s}}{(1+s^2)(s+1)}+\frac{1}{(s+1)}\\ \end{array}$

Using partial fraction we can write

$\begin{array}{c}x\left(s\right)=-e^{-\mathrm{Ï€}s}\left[\frac{1-s}{2(s^2+1)}+\frac{1}{2(s+1)}\right]+\frac{1}{(s+1)}\left(\right)\\ =-e^{-\mathrm{Ï€}s}\left[\frac{1}{2(s^2+1)}-\frac{s}{2(s^2+1)}+\frac{1}{2(s+1)}\right]+\frac{1}{(s+1)}\end{array}$

Taking inverse Laplace transform we get

$\begin{array}{c}x\left(t\right)=\left[\frac{-1sin(t-\mathrm{Ï€})}{2}+\frac{1cos(t-\mathrm{Ï€})}{2}-\frac{1e^{-(t-\mathrm{Ï€})}}{2}\right]u(t-\mathrm{Ï€})+e^{-t}\\ =[\frac{sint}{2}-\frac{cost}{2}-\frac{1e^{-(t-\mathrm{Ï€})}}{2}]u(t-\mathrm{Ï€})+e^{-t}\end{array}$

Hence

$\begin{array}{l}x\left(t\right)=\left[\frac{sint}{2}-\frac{cost}{2}-\frac{1e^{-(t-\mathrm{Ï€})}}{2}\right]u(t-\mathrm{Ï€})+e^{-t},y\left(t\right)=\left[\frac{cost}{2}+\frac{sint}{2}+1-\frac{1e^{-(t-\mathrm{Ï€})}}{2}\right]u(t-\mathrm{Ï€})+e^{-t}\\ \end{array}$

The final solution is

$\begin{array}{l}x\left(t\right)=\left[\frac{sint}{2}-\frac{cost}{2}-\frac{1e^{-(t-\mathrm{Ï€})}}{2}\right]u(t-\mathrm{Ï€})+e^{-t},y\left(t\right)=\left[\frac{cost}{2}+\frac{sint}{2}+1-\frac{1e^{-(t-\mathrm{Ï€})}}{2}\right]u(t-\mathrm{Ï€})+e^{-t}\\ \end{array}$