Solving Non Homogeneous Differential Equation, help. My task is to solve the non homogeneous system:
$x_1'=x_2+2t$
$x_2'=-x_1+2x_2-3$
with $F(t)=[2t,-3]$.
It can be rewritten as:
$x'=P(t)x+F(t)$
I know that to solve this I will first solve for the homogeneous case. Solving the homogeneous case using the concept of eigenvalue and eigenvector I got $c_1(1,1)e^t$ and $c_2(1,1)e^t$. So the general solution is $c_1(1,1)e^t+c_2(1,1)e^t$. But I am in doubt,my question is am I correct? And if I am, what shall I do next to solve the non homogeneous case? 
A detailed answer is highly appreciated.  
 A: I will give you a short answer which shows the method but leaves out the calculations:
We differentiate the second equation to obtain:
$$x_2''=-x_1'+2x_2'$$
Now substituting $x_1'$ from the first equation:
$$x_2''-2x_2'+x_2+2t=0$$
Which has general solution ($C_1,C_2$ arbitrary constants):
$$x_2(t)=C_1 e^t+C_2t e^t-2t-4$$
Now we find, using the first equation again:
$$x_1(t)=C_1e^t+C_2(t-1)e^t-4t+C_3$$
A: consider the differential equation $$\frac{dx}{dt} = Ax. \tag 1  $$
we will use the facts 
(a) if $u$ is an eigenvector corresponding to the eigenvalue $\lambda,$ then  $x = e^{\lambda t}u$ is a solution of $(1).$
(b) if $v$ is a generalized eigenvector satisfying $Av = \lambda v + u,$ then $x = (tu+v)e^{\lambda t}$ is a solution of $(1).$
in your problem $$A = \pmatrix{0&1\\-1&2}, \lambda = 1,1, u = \pmatrix{1\\1}, v = \pmatrix{0\\1}.$$  
so a solution of $(1)$ is $$x = C\pmatrix{1\\1}e^t + D\pmatrix{t\\t+1}e^t \text{ where $C,D$ are arbitrary constants.}$$
i solved only the homogeneous problem. i am assuming you know how to find a particular solution of the non homogeneous problem.  
