How can i solve this System of first-order differential Equations? My Problem is this given System of differential Equations:
$$\dot{x}=8x+18y$$ $$\dot{y}=-3x-7y$$
I am looking for a gerenal solution.
My Approach was: i can see this is a System of linear and ordinary differential equations. Both are of first-order, because the highest derivative is the first. But now i am stuck, i have no idea how to solve it. A Transformation into a Matrix should lead to this expression: $$\overrightarrow{y}=\left(
   \begin{array}{cc}
     8 & 18 \\
     -3 & -7
   \end{array}
\right)\cdot x$$
or is this correct: $$\overrightarrow{x}=\left(
   \begin{array}{cc}
     8 & 18 \\
     -3 & -7
   \end{array}
\right)\cdot y\text{ ?}$$
But i don't know how to determine the solution, from this point on.
 A: I'm going to rename your variables.  Instead of $x$ and $y$, I will use $x_1$ and $x_2$ (respectively).
Now, let's look at the system:
$$\begin{cases}
\dot x_1 = 8x_1+18x_2\\
\dot x_2 = -3x_1 -7x_2
\end{cases}$$
To change this into matrix form, we rewrite as $\dot {\vec x} = \mathbf A \vec x$, where $\mathbf A$ is a matrix.
This looks like:
$$\underbrace{\pmatrix{\dot x_1 \\ \dot x_2}}_{\large{\dot {\vec x}}} = \underbrace{\pmatrix{8 & 18 \\ -3 & -7}}_{\large{\mathbf A}}\underbrace{\pmatrix{x_1\\x_2}}_{\large{\vec x}}$$
To solve the system, we find the eigenvalues of the matrix.  These are $r_1 = 2$ and $r_2 = -1$.  Two corresponding eigenvectors are $\vec \xi_1 =\pmatrix{3 \\ -1}$ and $\vec \xi_2 =\pmatrix{2 \\ -1}$, respectively.
We now plug these into the equation:
$$\vec{x} = c_1e^{r_1t}\vec{\xi_1}+c_2e^{r_2t}\vec{\xi_2}$$
This yields:
$$\vec{x} = c_1e^{2t}\pmatrix{3 \\ -1}+c_2e^{-t}\pmatrix{2 \\ -1}$$
So, your individual solutions are:
$$x_1 = 3c_1e^{2t} + 2c_2e^{-t}\\
x_2 = -c_1e^{2t} -c_2e^{-t}$$
A: I think you want this matrix:
$$\left(
   \begin{array}{cc}
      \dot{x} \\
     \dot{y}
   \end{array}\right)=\left(
   \begin{array}{cc}
     8 & 18 \\
     -3 & -7
   \end{array}
\right)\cdot \left(
   \begin{array}{cc}
      x \\
     y
   \end{array}\right),$$
and you then diagonalize the coupling matrix to get decoupled equations.
Note, you have something like this:
$$\dot{X}=M\cdot X,$$
where: $$X=\left(
   \begin{array}{cc}
      x \\
     y
   \end{array}\right),\,\dot{X}=\left(
   \begin{array}{cc}
      \dot{x} \\
     \dot{y}
   \end{array}\right).$$
Then if the matrix $M$ is diagonalizable (this one is) you can write it as: $$M=S\cdot D \cdot S^{-1},$$
where $D$ is a diagonal matrix. You can then manipulate the differential equation as follows:
$$\dot{X}=S\cdot D \cdot S^{-1}\cdot X,$$
$$S^{-1}\cdot\dot{X}=D \cdot S^{-1}\cdot X,$$
$$\dot{U}=D \cdot U,$$
where $U=S^{-1}\cdot X$. This then gives you two decoupled differential equations to solve:
$$\dot{U_1}=D_1U_1.$$
$$\dot{U_2}=D_2U_2.$$
