<s>[INST] System of differential equations: why does it solve it the example? consider the system with conditions: $\quad y_1(0) = 1,\quad y_2(0) = -1,\quad y_3(0) = 0,\quad y_4(0) = 2$
$\begin{align}
&{y_1}'= y_1 + 6\,y_2\\
&{y_2}'= y_2 + 6\,y_3\\
&{y_3}'= y_3 + 6\,y_4\\
&{y_4}'= y_4
\end{align}$
what can be written as matrix: $A = \left(\begin{array}{cccc} 1 & 6 & 0 & 0\\ 0 & 1 & 6 & 0\\ 0 & 0 & 1 & 6\\ 0 & 0 & 0 & 1 \end{array}\right)$
Thus the solution: $\exp(A\,x)\,y_0 = e^{x}\,\underbrace{\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)}_{B} \,\left(\begin{array}{c}y_1(0)\\y_2(0)\\y_3(0)\\y_4(0)\end{array}\right) = B\,\left(\begin{array}{c}1\\-1\\0\\2\end{array}\right) $
Now to my question: how does it solve the system above? May sound stupid, because I understand solving something without knowing what. For example: what's $y_1, y_2,y_3,y_4$ in the solution matrix? The 1st, 2nd, 3rd, 4th row or entry? But even then I can't put it together.
Also: How is the starting condition vector built?  What would be the difference between $y_1(0) = 1$ and $y_1(1) = 1$
May be essential questions but these Systems are really kicking me off.
 [/INST] We are given the system
$$\begin{align}
&{y_1}'= y_1 + 6\,y_2 = 1\, y_1 + 6\, y_2 + 0 \,y_3 + 0\, y_4\\
&{y_2}'= y_2 + 6\,y_3 = 0 \,y_1 +1\, y_2 + 6\, y_3 + 0\, y_4\\
&{y_3}'= y_3 + 6\,y_4 = 0\, y_1 + 0\, y_2 + 1\, y_3 + 6\, y_4\\
&{y_4}'= y_4 = 0 \,y_1 + 0 \,y_2 + 0\, y_3 + 1\, y_4
\end{align}$$
This can be compactly written as
$$y' = A y =  \left(\begin{array}{cccc} 1 & 6 & 0 & 0\\ 0 & 1 & 6 & 0\\ 0 & 0 & 1 & 6\\ 0 & 0 & 0 & 1 \end{array}\right) \begin{pmatrix} y_1\\y_2\\y_3\\y_4 \end{pmatrix}$$
We now solve for the matrix exponential or matrix exponential (note that there are many approaches to solve a system, like eigenvalues/eigenvectors, diagonalization (when possible), Putzer's Algorithm...)
$$e^{Ax} = e^x\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)$$
Once we have the matrix exponential and are given initial conditions (they could have chosen anything), we can write the solution as
$$y(x) = e^{Ax} y_0 = e^x\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c}1\\-1\\0\\2\end{array}\right) = \left(
\begin{array}{c}
 72 e^x x^3-6 e^x x+e^x \\
 36 e^x x^2-e^x \\
 12 e^x x \\
 2 e^x \\
\end{array}
\right)$$
This means
$$\begin{align} y_1(x) &= 72 e^x x^3-6 e^x x+e^x\\ y_2(x) &= 36 e^x x^2-e^x \\ y_3(x) &= 12 e^x x \\y_4(x) &=  2 e^x \end{align}$$
We can verify this solution, for example
$$y_1' = y_1 + 6 y_2$$
Calculating the LHS
$$y_1' = 72 e^x x^3+216 e^x x^2-6 e^x x-5 e^x$$
Calculating the RHS
$$y_1 + 6 y_2 = 72 e^x x^3-6 e^x x+e^x + 6(36 e^x x^2-e^x) = 72 e^x x^3+216 e^x x^2-6 e^x x-5 e^x$$
The LHS $= $ RHS, that checks out and this looks like a valid solition - as will the three other results.
</s>