Question

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.

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

Short answer

The solution is

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

Step-by-step solution

Given information

The differential equations are given as:

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

Apply the Laplace transform

Given initial value equations are,

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

Taking Laplace transform of equation first we get

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

Taking Laplace transform of equation second we get

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

Putting equation fourth into third we get

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

Using partial fraction we can write as

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

Taking inverse Laplace transform we get

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

Use partial fraction

From equation fourth

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

Using partial fraction we can write

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

Taking inverse Laplace transform we get

$y\left(t\right)=\frac{e^t}{2}-\frac{e^{-t}}{2}-\left[\frac{{\displaystyle e^{(t-2)}}}{{\displaystyle 2}}-\frac{{\displaystyle e^{-(t-2)}}}{{\displaystyle 2}}\right]u(t-2)$

Hence

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

Conclusion

The final solution is

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