How can I solve this system of differential equations?
$\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ K_1 & K_2 & K_3 & K_4\\ T_1 &T_2 & T_3 & T_4 \\ \end{array} \right)*\left( \begin{array}{c} a' \\ b' \\ c' \\ d' \\ \end{array} \right)+\left( \begin{array}{cccc} A_1 & A_2 & 0 & 0 \\ B_1 & B_2 & B_3 & B_4 \\ 0 & 0 & C_3 & C_4 \\ 0 & 0 & D_4 & D_4 \\ \end{array} \right)*\left( \begin{array}{c} a \\ b \\ c \\ d \\ \end{array} \right)=\left( \begin{array}{c} E_1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right)$
$ K, T, A, B, C, D, E$ are constants.
Update: I edit the differential equations, it was my mistake
As written, your system has the form
$$A\mathbb y'=\mathbb b$$ or
$$\mathbb y'=A^{-1}\mathbb b,$$ trivially solved by integration as
$$\mathbb y(t)=A^{-1}\mathbb b\,t+\mathbb y_0.$$
Update: (after OP fixing the question)
The system has the form $$A\mathbb y'+B\mathbb y=\mathbb b$$
but due to the special structure of $A$, the first two equations are ordinary, so that $a,b$ can be expressed linearly in terms of $c,d$. Before adressing the differential part, eliminate $a,b$ and write the remaining two equations in the form
$$Pz'+Qz=r$$ where $\mathbb z=(c,d)$ and $P,Q$ are $2\times2$ matrices, or
$$\mathbb z'+P^{-1}Q\mathbb z=P^{-1}\mathbb r.$$
Now, the general solution of the homogeneous equation is
$$\mathbb z=C e^{-P^{-1}Qt}$$ where the exponential of the matrix can be obtained from its Eigenvalues.
A particular solution of the non-homogeneous equation is constant, given by
$$\mathbb z=Q^{-1}\mathbb r.$$
You combine these results to get the complete solution and plug the initial conditions to determine the matrix $C$.