How to resolve this differential equation with Laplace transform?

Question

$$\mathcal{L}\left\{ f'(t)\right\} =s\mathcal{L}\left\{ f(t)\right\} -f(0)$$

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$$\begin{cases} x' +x - y' = -t \\ x' + y' + y = 1 \end{cases}$$,

$$X=\mathcal{L}\left\{ x\right\}, \ Y=\mathcal{L}\left\{ y\right\}$$

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$$\begin{cases} sX-x(0) +X -sY+y(0) = \int_{0}^{+\infty}-te^{-st}\,dt \\ sX-x(0)+ sY-y(0)+ Y = \int_{0}^{+\infty}1\cdot e^{-st}\,dt \end{cases}$$

$$\begin{cases} sX-x(0) +X -sY+y(0) =-\frac{1}{s^2} \\ sX-x(0)+ sY-y(0)+ Y = \frac 1 s \end{cases}$$

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$$\begin{cases} (s+1)X -sY =-\frac{1}{s^2}+x(0)-y(0) \\ sX+(s+1)Y= \frac 1 s+x(0)+y(0) \end{cases}$$

$$\left(\begin{array}{cc}s+1 &-s\\s&s+1\end{array}\right)\left(\begin{array}{cc}X\\Y\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{s^2}+x(0)-y(0)\\\frac 1 s+x(0)+y(0) \end{array}\right)$$

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$$\left(\begin{array}{cc}s+1 &-s\\s&s+1\end{array}\right)^{-1} = \frac{1}{(s+1)^2+s^2}\left(\begin{array}{cc} s+1&s\\-s&s+1\end{array}\right)$$

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$$\left(\begin{array}{cc}X\\Y\end{array}\right)=\frac{1}{(s+1)^2+s^2}\left(\begin{array}{cc}-\frac{s+1}{s^2}+(s+1)(x(0)-y(0))+1+s(x(0)+y(0))\\\frac 1 s-s(x(0)-y(0))+\frac{s+1}{s}+(s+1)(x(0)+y(0)) \end{array}\right)$$

Answer 1

To compare against another solution method: Consider $z=x+iy$, then, using undetermined coefficients, \begin{align} (1+i)z'+z=i-t&\implies z = Ae^{-(1-i)t/2}+Bt+C,\\~\\ (1+i)B+Bt+C=i-t&\implies B=-1, C=1+2i\\~\\ z(0)=A+C&\implies A=z(0)-1-2i \end{align}

All together $$ z(t)=(z(0)-1-2i)e^{-(1-i)t/2}-t+1+2i, $$ so that in real components $$ x(t)=e^{-t/2}[(x(0)-1)\cos(t/2)-(y(0)-2)\sin(t/2)]-t+1\\ y(t)=e^{-t/2}[(x(0)-1)\sin(t/2)+(y(0)-2)\cos(t/2)]+2 $$

Answer 2

$${\cal L}^{-1} \left[ -\frac{s+1}{s^2}+(s+1)(x(0)-y(0))+1+s(x(0)+y(0)) \right] = x(0)\delta(t) - y(0)\delta(t) + \delta(t) + 2x(0)\delta^{'}(t) - t - 1 $$

$${\cal L}^{-1} \left [ \frac1 s-s(x(0)-y(0))+\frac{s+1}{s}+(s+1)(x(0)+y(0)) \right] = x(0)\delta(t) + y(0)\delta(t) + \delta(t) + 2y(0)\delta^{'}(t) +1 $$

so:

$$\left(\begin{array}{cc}X\\Y\end{array}\right)= e^{-\frac{t}{2}}\sin(\frac{t}{2}) \left(\begin{array}{cc}x(0)\delta(t) - y(0)\delta(t) + \delta(t) + 2x(0)\delta^{'}(t) - t - 1\\x(0)\delta(t) + y(0)\delta(t) + \delta(t) + 2y(0)\delta^{'}(t) +1\end{array}\right)$$

finally:

$$X(t) = e^{-\frac{t}{2}}\sin(\frac{t}{2}) \left( x(0)\delta(t) - y(0)\delta(t) + \delta(t) + 2x(0)\delta^{'}(t) - t - 1 \right)$$

$$Y(t) = e^{-\frac{t}{2}}\sin(\frac{t}{2}) \left( x(0)\delta(t) + y(0)\delta(t) + \delta(t) + 2y(0)\delta^{'}(t) \right)$$

and if I have a $y(0)$ and $x(0)$ i put to $Y(t)$ and $X(t)$ ? Or be considered as permanent for example $x(0) = A$ and $y(0)=B$.

$$X(t) = e^{-\frac{t}{2}}\sin(\frac{t}{2}) \left( A\delta(t) - B\delta(t) + \delta(t) + 2A\delta^{'}(t) - t - 1 \right)$$

$$Y(t) = e^{-\frac{t}{2}}\sin(\frac{t}{2}) \left( A\delta(t) + B\delta(t) + \delta(t) + 2B\delta^{'}(t) \right)$$