Finding y(t) as the solution of x' = Ax If I have a matrix A = \begin{bmatrix}2&0\\0&3\end{bmatrix} and I am trying to find the solution y(t) to x' = Ax (Solution: x(t) = x(t), y(t)), how should I go about it?
I have an initial condition that x(1) = \begin{bmatrix}123456789\\5\end{bmatrix}
Thanks.
 A: Let $\dot{x}$ and $\dot{y}$ be the derivatives of $x$ and $y$ with respect to $t$
You wish to solve the following:
$$\begin{bmatrix}2&0\\0&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix}$$
Note this system is decoupled and so you can solve each differential equation separately.  Specifically $\dot{x}=2x$ and $\dot{y}=3y$.
Taking the first equation:
$$\begin{align}\dot{x}&=2x\\\dot{x}-2x&=0 \end{align}$$
This is first order linear with constant coefficients and so can be solved with an integrating factor.  The integrating factor is $e^{\int-2\;dt}=e^{-2t}$
This essentially transforms the LHS in to the derivative of a product.
$$\begin{align}e^{-2t}\dot{x}-2xe^{-2t}&=0\\\frac{d}{dt}\left\{e^{-2t}x\right\}& =0\\e^{-2t}x& = C_1\\x&=C_1e^{2t}\end{align}$$
The second differential equation follows a similar path leading to a solution of $y=C_2e^{3t}$
You are given the state of the system at $t=1$ and so can find the values of $C_1$ and $C_2$ by substitution.  Namely:
$$123456789=C_1e^{2}\implies C_1=123456789e^{-2}\\5=C_2e^{3}\implies C_2=5e^{-3}$$
Hope this helps.  
