find all solution to the equation $y'=Ay+b(x)$ for given $A$ and $b(x)$ I am asked to find all solution to the equation:
$$y'= \left( \begin{array}{cc}
13 & 12 \\
12 & 13 \end{array} \right)  y+ \left(\begin{array}{c}
x\\
0 \end{array} \right)$$
No initial condition is specified. 
My working so far: If we write the DE as $y'=Ay+b(x)$ then a solution to the homogeneous DE $y'=Ay$ is given by:
$$y_h=e^{A(x-x_0)}y_0$$
where $y_0$ is a vector containing the arbitrary initial conditions. Because no initial conditions are specified could I just assume that $x_0=0$? I then probably need to use variation of parameters to find a general solution but I am unsure how to do this for a system of linear differential equations. Also, how do I calculate $e^{A(x-x_0)}$, is $A$ diagonalizable? If so I could calculate it. My linear algebra is not too great so I got stuck here too. Any help would be appreciated. Thanks!
 A: Do you know how to find the Eigenvalues and Eigenvectors of a matrix?
To find the eigenvalues, you solve for the roots of the characteristic polynomial using $(\mathbf{A} - \lambda \mathbf{I}) = 0$?
For your problem, you have 
$$ \mathbf{A} = \begin{bmatrix}
 13 & 12\\ 
 12 & 13 
\end{bmatrix} $$
We form $(\mathbf{A} - \lambda \mathbf{I}) = 0$, so
$$ (\mathbf{A} - \lambda \mathbf{I}) = \begin{bmatrix}
 13- \lambda & 12\\ 
 12 & 13 - \lambda
\end{bmatrix} = 0$$
$$(13 - \lambda)^{2} - 144 = 0, \text{so}, \lambda_{1,2} = 1, 25$$
To find the corresponding eigenvectors, you substitute each distinct (if they are not distinct, other approaches are needed) eigenvalue into and by solving $(\mathbf{A} - \lambda_i \mathbf{I})\mathbf{x} = 0$.
For this example, you would get:
$$\lambda_1 = 1, v1 = (1, 1)$$
$$\lambda_2 = 25, v2 = (-1, 1)$$
You could also approach this using the Jordan Normal Form and many other ways too.
Can you take it from here?
Regards
A: Take $x_0=0$ and $y_0=(c_1,c_2)^t\in\mathbb{R}^2$. As $A$ is real and symmetric, is diagonalizable in $\mathbb{R}$.  
