Inhomogenous Differential system This is the system,
$$ y'_1 =y_1+y_2+1 $$
$$ y'_2= -y_1+y_2+1 $$
initial value problem which fulfill:
$$y_1(0)=1$$
$$y_2(0)=-1$$
value to find
$$y_1(π)= \text{?}$$
 A: Use Laplace transform:
$$
\begin{cases}
y'_1(t)=y_1(t)+y_2(t)+1\\
y'_2(t)=y_2(t)-y_1(t)+1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\mathcal{L}_t\left[y'_1(t)\right]_{(s)}=\mathcal{L}_t\left[y_1(t)+y_2(t)+1\right]_{(s)}\\
\mathcal{L}_t\left[y'_2(t)\right]_{(s)}=\mathcal{L}_t\left[y_2(t)-y_1(t)+1\right]_{(s)}
\end{cases}\Longleftrightarrow
$$

Use:


*

*$$\mathcal{L}_t\left[1\right]_{(s)}=\frac{1}{s}$$

*$$\mathcal{L}_t\left[y_n(t)\right]_{(s)}=\text{Y}_n(s)$$

*$$\mathcal{L}_t\left[y'_n(t)\right]_{(s)}=s\text{Y}_n(s)-y_n(0)$$


$$
\begin{cases}
s\text{Y}_1(s)-y_1(0)=\text{Y}_1(s)+\text{Y}_2(s)+\frac{1}{s}\\
s\text{Y}_2(s)-y_2(0)=\text{Y}_2(s)-\text{Y}_1(s)+\frac{1}{s}
\end{cases}\Longleftrightarrow
$$

Use the initial conditions $y_1(0)=1$ and $y_2(0)=-1$:

$$
\begin{cases}
s\text{Y}_1(s)-1=\text{Y}_1(s)+\text{Y}_2(s)+\frac{1}{s}\\
s\text{Y}_2(s)+1=\text{Y}_2(s)-\text{Y}_1(s)+\frac{1}{s}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
s\text{Y}_1(s)-\text{Y}_1(s)=\text{Y}_2(s)+\frac{1}{s}+1\\
s\text{Y}_2(s)-\text{Y}_2(s)=\frac{1}{s}-1-\text{Y}_1(s)
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\text{Y}_1(s)\left[s-1\right]=\text{Y}_2(s)+\frac{1}{s}+1\\
\text{Y}_2(s)\left[s-1\right]=\frac{1}{s}-1-\text{Y}_1(s)
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\text{Y}_1(s)=\frac{\text{Y}_2(s)+\frac{1}{s}+1}{s-1}\\
\text{Y}_2(s)=\frac{\frac{1}{s}-1-\text{Y}_1(s)}{s-1}
\end{cases}
$$
Now, using substitution:


*

*$$\text{Y}_1(s)=\frac{s-1}{2+s(s-2)}$$

*$$\text{Y}_2(s)=-\frac{1}{s}-\frac{1}{2+s(s-2)}$$


With inverse Laplace transform:


*

*$$y_1(t)=e^t\cos(t)$$

*$$y_2(t)=-1-e^t\sin(t)$$


So, for $y_1(\pi)$:
$$y_1(t)=e^t\cos(t)\to y_1(\pi)=e^{\pi}\cos(\pi)=e^{\pi}\cdot-1=-e^{\pi}\approx-23.1407$$
A: Another solution, using linear algebra.
Consider the matrix $A = \pmatrix{ 1 & 1 \\ -1 & 1}$ and $b(t) = \pmatrix{1 \\ 1}$. Let $Y(t) = \pmatrix{y_1(t) \\ y_2(t)}$.
The problem is equivalent to solving
$$ Y'(t) = AY(t) + b(t).$$
Remark that the characteristic polynomial of $A$ is $X^2 - 2X + 2 = (X-1+i)(X-1-i)$, meaning $A$ is diagonalizable. Two associated eigenvectors are $v_1 = \pmatrix{1 \\ -i}$ and $v_2 = \pmatrix{1\\i}$.
Let $P = \pmatrix{1 & 1 \\ -i & i}$ such that $P^{-1}AP = D = \pmatrix{1-i & 0 \\ 0 & 1+i }$, and $Y = PX$. One has
$$ Y' = AY + b \Longleftrightarrow PX' = APX + b$$
hence
$$X' = P^{-1}APYX + P^{-1}b = DX + P^{-1}b$$
One has $P^{-1} = \pmatrix{\frac{1}{2} & \frac{i}{2} \\ \frac{1}{2} & -\frac{i}{2}}$ hence $P^{-1}b(t) = \frac{1}{2}\pmatrix{1+i \\ 1-i}$. The system is then equivalent to
$$ \cases{x_1' = (1-i)x_1 + \frac{1}{2}(1+i) \\ x_2' = (1+i)x_2 + \frac{1}{2}(1-i)} $$
which solutions are
$$ \cases{x_1(t) = c_1 e^{(1-i)t}-\frac{i}{2} \\ x_2(t) = c_2 e^{(1+i)t}+\frac{i}{2} }, \qquad (c_1, c_2) \in \mathbb{R}^2.$$
Initial conditions lead to 
$$ c_1 = c_2 = \frac{1}{2}. $$
Now, using $Y=PX$, one finally has
$$ \bbox[lightgreen,5px,border:2px solid green]{\cases{y_1(t) = e^t\left(\dfrac{e^{it} + e^{-it}}{2}\right) = e^t \cos(t) \\ y_2(t)  = ie^t\left(\dfrac{e^{it} - e^{-it}}{2}\right) - 1 = -e^t \sin(t) - 1}}. $$
One can directly deduce that $\bbox[lightgreen,5px,border:2px solid green]{y_1(\pi) = -e^{\pi}}$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets
  $\ds{%
\,\mathsf{A} \equiv 
\pars{\begin{array}{rr}
\ds{0} & \ds{-\ic}
\\
\ds{\ic} & \ds{0}
\end{array}}\
\mbox{and}\
\,\mathsf{A}_{0} \equiv 
\pars{\begin{array}{rr}
\ds{1} & \ds{0}
\\
\ds{0} & \ds{1}
\end{array}}}$. Note that $\ds{\,\mathsf{A}^{2} = \,\mathsf{A}_{0}}$.

\begin{align}
&\pars{\totald{}{x} - \,\mathsf{A}_{0} - \ic\,\mathsf{A}}
{\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x} + 1} = {0 \choose 0}\
\imp\
{\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x} + 1} = 
\exp\pars{\bracks{\,\mathsf{A}_{0} + \ic\,\mathsf{A}}x}{1 \choose 0}
\end{align}

\begin{align}
{\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x} + 1} & = 
\exp\pars{x}\exp\pars{\ic\,\mathsf{A}x}{1 \choose 0} =
\exp\pars{x}\bracks{\cos\pars{x} + \ic\,\mathsf{A}\sin\pars{x}}{1 \choose 0}
\\[4mm] & =
\exp\pars{x}
\pars{\begin{array}{rr}
\ds{\cos\pars{x}} & \ds{\sin\pars{x}}
\\[2mm]
\ds{-\sin\pars{x}} & \ds{\cos\pars{x}}
\end{array}}{1 \choose 0}
\end{align}

$$
\,\mathrm{y}_{1}\pars{x} =
\expo{x}\cos\pars{x}\quad\imp\quad\,
\color{#f00}{\mathrm{y}_{1}\pars{\pi}} = \color{#f00}{-\expo{\pi}}
$$
