Solving inhomogenous continuous-time system with non-diagonalisable system matrix I have an exercise where i have to find the general solution to this problem:
$$
X'=\left(
\begin{matrix}
2&-1\\
 4&-2
\end{matrix}
\right)X + 
\begin{pmatrix}
2\\1
\end {pmatrix}. 
$$
Obviously the system matrix has linearly dependent vectors, which means that it can't be diagonalsed. Another method I tried is to use an expansion and using it I got a homogeneous solution:
$$
X_h = e^{tA} \begin{pmatrix}C_1\\C_2\end{pmatrix}
$$
Where
$$
e^{tA} = I + tA + \frac{t^2A^2}{2!} + ... , where A^2 = 0
$$
The homogeneous solution is then
$$
\begin{pmatrix}
1+2t&-t\\
 4t&1-2
\end{pmatrix} *
\begin{pmatrix}
C_1\\C_2
\end{pmatrix}
$$ 
The problem is I dont know how to get the particular solution. I tried using 
$$
X_p = (sI - A)^{-1}\pmatrix{2\\1}e^{st}
$$
Where s in this case is zero, but the matrix -A is not invertible. Can anyone help me?
 A: If $A^2=0$ you can find easily $e^{At}=I+tA$ by the Taylor expansion. Now the solution is given by the formula
$$
X(t)=e^{At}X(0)+e^{At}\int_0^te^{-As}\left[\matrix{2\\1}\right]\,ds.\tag1
$$
P.S. To get $(1)$ use the integrating factor $e^{-At}$ for the system
$$
\underbrace{e^{-At}(X'-AX)}_{(e^{-At}X(t))'}=e^{-At}\left[\matrix{2\\1}\right]\quad\Leftrightarrow\quad
e^{-At}X(t)-X(0)=\int_0^t e^{-As}\left[\matrix{2\\1}\right]\,ds.
$$

P.P.S. How the integrating factor works


*

*We collect all terms that depend on $X$ on the left hand side and check if the derivative has the unit coefficient
$$
X'(t)-AX(t)=f(t).
$$

*Now we would like to get a complete derivative in the left hand side after multiplying both sides of the equation by a proper chosen function, let call it $\Phi(t)$
$$
\color{red}{\Phi(t)}(X'-AX)=\color{red}{\Phi(t)}f(t)\qquad\Leftrightarrow\qquad
\Phi(t)\cdot X'(t)+\underbrace{\Phi(t)(-A)}_{=\Phi'(t)}X(t)=\Phi(t)f(t)
$$

*We want the left hand side to be the complete derivative of $\Phi\cdot X$, which leaves us almost no other choice for $\Phi$ than $\Phi(t)=e^{-At}$. With this choice we have
$$
(e^{-At}X(t))'=e^{-At}f(t).
$$

*Now integrate both sides from $0$ (or any other initial value for $t$) to $t$. Because the variable $t$ is occupied in the integral limit, we change the integration variable name to your favorite name (I like $s$) to avoid confusions
$$
\int_0^t(e^{-As}X(s))'\,ds=\int_0^t e^{-As}f(s)\,ds\qquad\Leftrightarrow\qquad
\underbrace{\left[e^{-As}X(s)\right]_0^t}_{=e^{-At}X(t)-I\cdot X(0)}=\int_0^t e^{-As}f(s)\,ds.
$$

*Move $X(0)$ to the right and multiply both sides by $e^{At}$ to get $(1)$.

