I've got stuck with this differential equation. I've to solve the problem using the exponential expansion.
$$ y'(x) = \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} y \text{ }\text{ }; y(1) = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} $$
Let's get $A^n$:
$$ A^2 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \\ \end{pmatrix} \text{ }\text{ } A^3 = \begin{pmatrix} 4 & 4 \\ 4 & 4 \\ \end{pmatrix} \text{ }\text{ } ... \text{ }\text{ } A^n = 2^{n-1} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$
We can do the exponential expasion like:
$$ e^{Ax} = \sum_{n=0}^{\infty} \frac{(Ax)^n}{n!} = \sum_{n=0}^{\infty} 2^{n-1} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \frac{x^n}{n!} = \frac {1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \frac {1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}e^{2x} $$
So that, I think the solution should be:
$$ y(x) = \frac {C}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}e^{2x} $$
But I think this is wrong because it looks like it's impossible to get the particular solution:
$$ y(1) = \frac {1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \begin{pmatrix} C_1 \\ C_2 \\ \end{pmatrix} e^2 = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} $$
As you can see, this is not possible to get a solution from this equation.
Does anyone know where I made a mistake? Thanks in advance.
The formula $$ A^n = 2^{n-1} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$ is not correct for $n=0$. $$ e^{Ax} = \sum_{n=0}^{\infty} \frac{(Ax)^n}{n!} = I+\sum_{n=1}^{\infty} 2^{n-1} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \frac{x^n}{n!} = I+ \frac {1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \sum_{n=1}^{\infty} \frac{(2x)^n}{n!} $$ $$ = I+\frac12\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}+ \frac {1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \sum_{n=1}^{\infty} \frac{(2x)^n}{n!} -\frac12\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$ $$ = I+\frac {1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}e^{2x}-\frac12\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$