The ODE $y''(x)=\sinh(x)-3y'(x)-2y(x)$ I am trying to solve the differential equation that is in the title as a System of first order ode.
My Approach:
$\frac{d}{dx} \left(\begin{array}{c} y \\ y' \end{array}\right)=\left(\begin{array}{c} y' \\ \sinh(x)-3y'-2y \end{array}\right)=$
$\left( \begin{array}{rrr}
0 & 1 \\ 
-2 & -3 \\
\end{array}\right)\left(\begin{array}{c} y \\ y' \end{array}\right)+\left(\begin{array}{c} 0 \\ \sinh(x) \end{array}\right)$
Then I calculate the characteristic polynomial of the coefficient matrix, which leads to the eigenvalues $\lambda_1=-2,\lambda_2=-1$. Let A denote the matrix above.
Calculating the matrix exponential, I get $e^{Ax}$=$\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\ 
-2e^x+2 & -e^x+2 \\
\end{array}\right)$
Now I am variating the parameters and get
$y(x)=\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\ 
-2e^x+2 & -e^x+2 \\
\end{array}\right)y_0+\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\ 
-2e^x+2 & -e^x+2 \\
\end{array}\right) \int_0^s  \left(\begin{array}{c} \sinh(x)(e^{-s}-1) \\ \sinh(x)(-e^{-s}+2) \end{array}\right)ds$ =
$y(x)=\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\ 
-2e^x+2 & -e^x+2 \\
\end{array}\right)y_0+\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\ 
-2e^x+2 & -e^x+2 \\
\end{array}\right)  \left(\begin{array}{c} \frac{1}{4}e^{-2x}(e^{2x}(2x+3)-2e^x-2e^{3x}+1+C_1) \\ \frac{-x}{2} -\frac{e^{-2x}}{4}+e^{-x}+e^{x} +C_2\end{array}\right)$
My Questions are: In the exercise description there was no value for $y_0$, is there a way to find the value for it?
Is my calculation correct (does it seem correct) or are there any mistakes?

Rewrite equation as
$$y''+3y'+2y=\frac{e^x}{2}-\frac{e^{-x}}{2}$$
Characteristic equation is
$$\lambda^2+3\lambda + 2=0$$
Then $\lambda_1=-2,\lambda_2=-1$. Solution of $y''+3y'+2y=0$ is
$$y_h=c_1e^{-2x}+c_2e^{-x}$$
For particular solution use method of undetermined coefficients.
$$y_p=Ae^x+Bxe^{-x}$$
We get
$$6Ae^x+Be^{-x}=\frac{e^x}{2}-\frac{e^{-x}}{2}$$
Then $A=\frac{1}{12}, B=-\frac{1}{2}$.
General solution is
$$y=y_h+y_p=c_1e^{-2x}+c_2e^{-x}+\frac{e^x}{12}-\frac{xe^{-x}}{2}.$$
