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.

$\begin{array}{ccc}{\mathbf{\text{x}}}^{\mathbf{\text{''}}}\mathbf{\text{=y+u(t-1);}}& \mathbf{\text{x(0)=1,}}& {\mathbf{\text{x}}}^{\mathbf{\text{'}}}\mathbf{\text{(0)=0}}\\ {\mathbf{\text{y}}}^{\mathbf{\text{''}}}\mathbf{\text{=x+1-u(t-1);}}& \mathbf{\text{y(0)=0,}}& {\mathbf{\text{y}}}^{\mathbf{\text{'}}}\mathbf{\text{(0)=0}}\end{array}$

The differential equations are given as:

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

Given initial value equations are,

$\begin{array}{ccc}{x}^{\text{'}\text{'}}=y+u(t-1);& x\left(0\right)=1,& x^{\text{'}}\left(0\right)=\mathrm{0....}\left(\text{1}\right)\\ y^{\text{'}\text{'}}=x+1-u(t-1);& y\left(0\right)=0,& y^{\text{'}}\left(0\right)=\mathrm{0.....}\left(\text{2}\right)\end{array}$

Taking Laplace transform of equation first we get

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

Taking Laplace transform of equation second we get

$\begin{array}{c}{s}^{2}y\left(s\right)-s{y}^{\text{'}}\left(0\right)-y\left(0\right)=x\left(s\right)+\frac{1}{s}-\frac{e^{-s}}{s}{s}^2\\ y\left(s\right)=x\left(s\right)+\frac{1}{s}-\frac{e^{-s}}{s}\\ x\left(s\right)=s^{2}y\left(s\right)-\frac{1}{s}+\frac{e^{-s}}{s}\mathrm{.....}\left(4\right)\end{array}$

Putting equation fourth into third we get

$\begin{array}{c}{s}^{4}y\left(s\right)-s+s{e}^{-s}-1=y\left(s\right)+\frac{e^{-s}}{s}\\ (s^4-1)y\left(s\right)=\frac{e^{-s}}{s}+s-s{e}^{-s}+1\\ y\left(s\right)=\frac{e^{-s}}{s(s^4-1)}+\frac{s-s{e}^{-s}+1}{(s^4-1)}\\ =\frac{(1-s^2)e^{-s}}{s(s^4-1)}+\frac{s+1}{(s^4-1)}\end{array}$

Taking partial fraction we get

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

Taking inverse Laplace transform we get

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

Similarly solving equation fourth and third we get

$x\left(s\right)=\frac{e^{-s}(s^2-1)}{s(s^4-1)}+\frac{1+s^3}{s(s^4-1)}$

Taking partial fraction we get

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

Taking inverse Laplace transform we get

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

Hence

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

The final solution is

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