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.

${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{2}$$\mathit{x}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$

${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{x}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{1}$$\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Since the initial conditions are given at $t=1$, we need to shift the argument to the zero, so let

$\begin{array}{l}{x}_1\left(t\right)=x(t+1)\\ y_1\left(t\right)=y(t+1)\end{array}$

Then $x_1\left(0\right)=x\left(1\right)=1$and $y_1\left(0\right)=y\left(1\right)=0$and

$\begin{array}{l}{x}_1^{\text{'}}\left(t\right)=x^{\text{'}}(t+1)(t+1)=x^{\text{'}}(t+1)\\ y_1^{\text{'}}\left(t\right)=y^{\text{'}}(t+1)(t+1)=y^{\text{'}}(t+1)\end{array}$

Then the given system becomes

$\{\begin{array}{l}{x}_1^{\text{'}}-2y_1=2\\ x_1^{\text{'}}+x_1-y_1^{\text{'}}={(t+1)}^2-2(t+1)-1\end{array}\Rightarrow \{\begin{array}{l}{x}_1^{\text{'}}-2y_1=2,x_1\left(0\right)=1\\ x_1^{\text{'}}+x_1-y_1^{\text{'}}=t^2+4t+2,y_1\left(0\right)=0\end{array}$

Applying the Laplace transform gives

$\{\begin{array}{l}L\{x_1^{\text{'}}-2y_1\}\left(s\right)=L\left\{2\right\}\left(s\right)\\ L\{x_1^{\text{'}}+x_1-y_1^{\text{'}}\}\left(s\right)=L\{t^2+4t+2\}\left(s\right)\end{array}$

Simplify equation, further as:

$\Rightarrow \{\begin{array}{l}s{X}_1\left(s\right)-x_1\left(0\right)-2Y_1\left(s\right)=\frac{2}{s}\\ s{X}_1\left(s\right)-x_1\left(0\right)+X_1\left(s\right)-s{Y}_1\left(s\right)-y_1\left(0\right)=\frac{2}{s^3}+\frac{4}{s^2}+\frac{2}{s}\end{array}$

$\Rightarrow \{\begin{array}{l}s{X}_1\left(s\right)-1+2Y_1\left(s\right)=\frac{2}{s}\\ s{X}_1\left(s\right)-1+X_1\left(s\right)-s{Y}_1\left(s\right)=\frac{2}{s^3}+\frac{4}{s^2}+\frac{2}{s}\end{array}$

$\Rightarrow \{\begin{array}{l}{X}_1\left(s\right)=\frac{2}{s^2}+\frac{2}{s}{Y}_1\left(s\right)+\frac{1}{s}\\ (2+\frac{2}{s}-s)Y_1\left(s\right)=\frac{2}{s^3}-\frac{1}{s}+\frac{2}{s^2}\end{array}$

Simplify equation$(2+\frac{2}{s}-s)Y_1\left(s\right)=\frac{2}{s^3}-\frac{1}{s}+\frac{2}{s^2}$, further as:

$\begin{array}{c}{Y}_1\left(s\right)=\frac{2-s^2+2s}{s^3}\\ =\frac{1}{s^2}\end{array}$

Substituting $\frac{1}{s^2}$ for $Y_1\left(s\right)$in equation (1) gives

$X_1\left(s\right)=\frac{2}{s^2}+\frac{2}{s^3}+\frac{1}{s}$

Applying inverse Laplace we get

$\begin{array}{l}{L}^{-1}\left\{Y_1\left(s\right)\right\}\left(t\right)=L^{-1}\left\{\frac{1}{s^2}\right\}\left(t\right)Þy_1\left(t\right)=t\\ L^{-1}\left\{X_1\left(s\right)\right\}\left(t\right)=L^{-1}\{\frac{2}{s^2}+\frac{2}{s^3}+\frac{1}{s}\}\left(t\right)\end{array}$

Take inverse Laplace transform we get;

$\begin{array}{c}{x}_1\left(t\right)=t^2+2t+1\\ ={\left(t+1\right)}^2\end{array}$

Now we find

$\begin{array}{l}x\left(t\right)=x_1(t-1)=t^2\\ y\left(t\right)=y_1(t-1)=t-1\end{array}$

Therefore

$x\left(t\right)=t^2,y\left(t\right)=t-1$