how resolve this difference equation used Laplace transform?

Question

Solve the equation or system of equations using the Laplace transform.

I need a solution to this task. Anyone could do this for me?

\begin{cases} x' +y' -x = 1 \\ x' + 2y' = 0 \end{cases}

$$x(0) = 0$$

$$ y(0) = 1$$

Answer 1

Let $$ X(s)=L\{x(t)\}, Y(s)=L\{y(t)\} $$ and then $$ L\{x'(t)\}=sX(s)-x(0)=sX(s), L\{y'(t)\}=sY(s)-y(0)=sY(s)-1. $$ Taking Laplace transform for these two equations gives $$ sX(s)+sY(s)-1-X(s)=\frac1s, sX(s)+2sY(s)-2=0$$ or $$ (s-1)X(s)+sY(s)=1+\frac1s, sX(s)+2sY(s)=-2.$$ Solving $X(s)$ and $Y(s)$ and taking inverse Laplace transform, you will get the answer. I omit the detail.