Solving system of linear differential equations by eigenvalues Using eigenvalues and eigenvectors solve system of differential equations:
$$x_1'=x_1+2x_2$$
$$x_2' = 2x_1+x_2$$
And find solution for the initial conditions: $x_1(0) = 1; x_2(0) = -1$
I tried to solve it, but I don't have right results, so I can't check my solution. I would like someone to write how he would solve it and what results would he get.
 A: i really need to get more reputation so i can comment. anyway heres my attempt. hope it helps.
you have 
$$\dot{x}=x+2y\\\dot{y}=2x+y$$
the systems linear so it can be rewritten as
$$\vec{\dot{x}}=A\vec{x}$$
$$\vec{\dot{x}}=\begin{bmatrix}
    1 & 2 \\
2 & 1
\end{bmatrix}\begin{bmatrix}
    x \\
y
\end{bmatrix}$$
calculating $det[A-\lambda I]$ gives $det\begin{bmatrix}
    1-\lambda & 2 \\
2 & 1-\lambda
\end{bmatrix}$
simplifing to $\lambda^{2}-2\lambda-3=0 \Rightarrow (\lambda-3)(\lambda+1)$
Calculate the eigan vectors now.
$\begin{bmatrix}
    1-3 & 2 \\
2 & 1-3
\end{bmatrix}\begin{bmatrix}
    \vec{v_{1}}\\
\vec{v_{2}}
\end{bmatrix}=0 \Rightarrow\begin{bmatrix}
   1\\
1
\end{bmatrix}$
and
$\begin{bmatrix}
    1--1 & 2 \\
2 & 1--1
\end{bmatrix}\begin{bmatrix}
    \vec{v_{1}}\\
\vec{v_{2}}
\end{bmatrix}=0 \Rightarrow\begin{bmatrix}
   1\\
-1
\end{bmatrix}$
(sorry if someone could check this last one for me that'd be great)
This gives the solution
$$\vec{x}=c_{1}e^{3t}\begin{bmatrix}1\\1\end{bmatrix}+c_{2}e^{-t}\begin{bmatrix}1\\-1\end{bmatrix}$$
substituting initial conditions give 
$$\vec{x(0)}=c_{1}\begin{bmatrix}1\\1\end{bmatrix}+c_{2}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}$$
implies $c_{1}=1 $ and $c_{2}=0$
giving final solution $$\vec{x}=e^{3t}\begin{bmatrix}1\\1\end{bmatrix}$$
A: The system matrix is
$$A=\begin{bmatrix}1 &2\\ 2& 1
\end{bmatrix}$$
The eigenvalues are $-1$ and $3$.
$$\Lambda=\begin{bmatrix}-1 & 0\\0& 3\end{bmatrix}$$
The eigenvector matrix is (normalized columns)
$$Q=\begin{bmatrix}-0.7071 &    0.7071\\
0.7071  &  0.7071\end{bmatrix}=\begin{bmatrix}q_1 &q_2\end{bmatrix}$$
This matrix is orthogonal.
Then $x=[x_1 \quad x_2]'$:
$$x(t)=e^{At}x(0)=Qe^{\Lambda t}Q'x(0)=q_1q_1'x(0)e^{-t}+q_2q_2'x(0)e^{3t}$$
A: The matrix A of coefficients has two eigenvalues, $3$ and $-1$, and the corresponding eigenvectors $$v_1=(1,1)\quad v_2(1,-1)$$ If you write your system like: $Av=v'$ where $v=(x_1(t),x_2(t))$ you find that $v_1(t)=e^{3t}v_1$ and $v_2(t)=e^{-t}v_2$ you see that (the) solution is given by $v=c_1v_1(t)+c_2v_2(t)$. Then you impose the initial conditions. Sorry for the poor english.
A: Look here and here, and following their notation,$$A = \begin{pmatrix}
1 & 2 \\
2 & 1 \\
\end{pmatrix},$$ $$\lambda_1 = 3 \text{ and } \lambda_2 = -1,$$ $$\text{& } \vec{\eta}_1 = \begin{pmatrix}
1 \\
1 \\
\end{pmatrix} \text{ and } \vec{\eta}_2 = \begin{pmatrix}
1 \\
-1 \\
\end{pmatrix}.$$Then, $\vec{x}(t)=c_1e^{3t}\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}+c_2e^{-t}\begin{pmatrix}
1 \\
-1 \\
\end{pmatrix} = \begin{pmatrix}
c_1e^{3t}+c_2e^{-t} \\
c_1e^{3t}-c_2e^{-t} \\
\end{pmatrix}.$
Therefore, $\begin{cases}
x_1=c_1e^{3t}+c_2e^{-t} \\
x_2=c_1e^{3t}-c_2e^{-t}.
\end{cases}$  
Don't forget to solve for $c_1$ and  $c_2$ as well.
