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?
 A: 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
$$y=y_h+y_p=c_1e^{-2x}+c_2e^{-x}+\frac{e^x}{12}-\frac{xe^{-x}}{2}.$$
A: There is no difference between this method and the classical one for linear ODE's with constant coefficient. The Eigenvalues are found by solving the characteristic polynomial $$\lambda(\lambda+3)+2$$ while the straight method yields the characteristic polynomial
$$\lambda^2+3\lambda+2.$$
You just get longer computation because you compute $y$ and $y'$ simultaneously.
Alternatively to variation of the coefficients, you can use the known rules for exponential RHS, or indeterminate coefficients based on an ansatz with $\sinh,\cosh$ times first degree polynomials (needed because one root is common with the time constant of an exponential).

The value of $y_0$ is a free parameter (there is a solution for any $y_0$), you can't compute it.
A: We could utilize Duhamel's Integral for this problem. First, I'll write the end goal so that given $y''(x)+a*y'(x)+b*y(x)=g(x)$:
$$y(x)=y_{h}(x)+\int_{0}^{x}u(x-s)*g(s)ds$$
Where $u(x,A,B)=y_{h}(x,c_{1},c_{2})$ is solved for constants $A,B$.
First we observe the characteristic polynomial $y_{c}(\lambda)$ such that:
$$y_{c}(\lambda)=0=\lambda^2+a\lambda+b \implies 0=\lambda^2+3\lambda+2\implies **\lambda \in \lbrace -2 , -1 \rbrace**$$
Each characteristic root is of multiplicity $1$ so our homogenous solution takes the form:
$$y_{h}(x)=c_{1}x^{0}e^{-2x}+c_{2}x^{0}e^{-1x} \implies **y_{h}(x)=c_1e^{-2x}+c_{2}e^{-x}**$$
Now we obtain $u(x)$ with solved constants:
$$u(0)=0=A+B \implies A=-B$$
$$u'(0)=1=-2A-B \implies A=-1 , B=1$$
$$**u(x)=-e^{-2x}+e^{-x}**$$
Now we can form the solution so that:
$$**y(x)=c_1e^{-2x}+c_2e^{-x}+\int^{x}_{0}(-e^{-2(x-s)}+e^{-(x-s)})*(sinh(s))ds**$$
The specifics on this solution can be further explored in Ordinary Differential Equations and Dynamical Systems by Teschl in chapter 3. Because I found it hard enough to learn how to form this solution so that I could be typing it here, I'll leave the actual integration as an exercise. :)
