IVP with challenging numbers. I've seen it evaluated by non trivial manipulations. Can someone complete it step by step with explanations? Solve the IVP: $X' = AX+f(t)$
$$\begin{align*}
A&=
\begin{bmatrix}6/7 & -15/14\\-5/7 & 37/14\end{bmatrix}
\\
X(0)&=
\begin{bmatrix}4\\-1\end{bmatrix}
\\
f(t)&= \begin{bmatrix}e^{2t}\\e^{-t}\end{bmatrix}
\\
\end{align*}
$$
 A: Let, in general, $$x'=Ax+f\tag 1$$ a system of $n$ coupled ODE with $x(t)$ and $f(t)$ two $n\times 1$ vectors and $A$ a $n\times n$ matrix of constants. One way to solve these $n$ coupled first order linear ODE is to diagonalize the coefficient matrix $A$ and thus decouple these equations. 
Suppose that $\lambda_1,\ldots,\lambda_n$ and $v_1,\ldots, v_n$ are distinct eigenvalues and associated eigenfvectors of $$Av=\lambda v$$ and $V=(v_1,\ldots,v_n)$ the matrix of eigenvectors; putting $x=Vz$ the system (1) can be written as
$$
Vz'=AVz+f
$$
that is $$z'(t)=V^{-1}AVz(t)+V^{-1}f(t)=\Lambda z(t)+\phi(t)$$
with $\Lambda=V^{-1}AV=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ and $\phi(t)=V^{-1}f(t)$.
The general solution of this system of equations
$$z'_k(t)=\lambda_k z_k(t)+\phi_k(t)\qquad k=1,\ldots,n $$
is
$$z_k(t)=c_k\operatorname{e}^{\lambda_k t}+\operatorname{e}^{\lambda_k t}\int\operatorname{e}^{-\lambda_k t}\phi_k(t)\operatorname{d}t\qquad k=1,\ldots,n $$
Substituting back to $x=Vz$, the original system of differential equations can be solved as
$$
x(t)=\sum_{k=1}^n c_kv_k\operatorname{e}^{\lambda_k t}+\sum_{k=1}^n v_k\operatorname{e}^{\lambda_k t}\int\operatorname{e}^{-\lambda_k t}\phi_k(t)\operatorname{d}t
$$
and the constant are determinde with the initial condition $x(0)=x_0$.
So, following these steps, you can solve your homework.
A: If solving by decoupling, first find eigenvalues and corresponding eigenvectors.
e1 = 1/2 ; e2 = 3
$$v1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$$ $$v2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$$
$$P = \begin{bmatrix} 3 & 1 \\ 1 & -2 \end{bmatrix}$$
$$P^-1 = \begin{bmatrix} 2/7 & 1/7 \\ 1/7 & -3/7 \end{bmatrix}$$
$$P^-1*f(t) = \begin{bmatrix} (2/7)e^{2t}+e^{-t}/7 \\ e^{2t}/7 - (3/7)e^{-t} \end{bmatrix}$$
$$\begin{bmatrix} y'1 \\ y'2 \end{bmatrix} = \begin{bmatrix} e1*y1 \\ e2*y2 \end{bmatrix} + \begin{bmatrix} (2/7)e^{2t}+e^{-t}/7 \\ e^{2t}/7 - (3/7)e^{-t} \end{bmatrix}$$
At this point you have two linear differential equations set up to solve.
You can solve them using an integrating factor or Laplace transforms if you like.
(1)  y'1 - (1/2)y1 = (2/7)e^{2t} + e^{-t}/7
(2)  y'2 - 3y2 = e^{2t}/7 - (3/7)e^{-t}
$$ X(t) = \begin{bmatrix} y1 \\ y2 \end{bmatrix} = \begin{bmatrix} (4/21)e^{2t}-(2/21)e^{-t}+c1e^{t/2} \\ -(1/7)e^{2t}+(3/28)e^{-t}+c2e^{3t} \end{bmatrix}$$
X(t) is your solution. All that's left to do is solve for your constants c1 and c2 using your initial condtions $$ X(0) = \begin{bmatrix} 4 \\ -1 \end{bmatrix}$$
c1 = 82/21  c2 = -27/28
Your final solution should look like this
$$ X(t) = \begin{bmatrix} (4/21)e^{2t}-(2/21)e^{-t}+(82/21)e^{t/2} \\ -(1/7)e^{2t}+(3/28)e^{-t}-(27/28)e^{3t} \end{bmatrix}$$
