In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

$\begin{array}{c}{\mathbf{\text{x}}}^{\mathbf{\text{'}}}{\mathbf{\text{-y}}}^{\mathbf{\text{'}}}\mathbf{\text{=(sint)u(t-π);\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x(0)=1,}}\\ {\mathbf{\text{x+y}}}^{\mathbf{\text{'}}}\mathbf{\text{=0;\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y(0)=1}}\end{array}$

The differential equations are given as:

$\begin{array}{l}{x}^{\text{'}}-y^{\text{'}}=\left(sint\right)u(t-\pi );x\left(0\right)=1,\\ x+y^{\text{'}}=0;y\left(0\right)=1\end{array}$

Given initial value equations are,

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

Taking Laplace transform of equation first we get

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

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

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

Simplify further as:

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

Using partial fraction we get

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

Taking inverse Laplace transform we get

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

From equation fourth

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

Using partial fraction we can write

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

Taking inverse Laplace transform we get

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

Hence

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

The final solution is

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