Marco2

There are several methods to solve the following system of equations.

(1)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_1' & = & x_2 - 2x_1 \\ x_2' & = & x_1 - 2x_2 \\ \end{eqnarray}

The first method involves combining the equations:

(2)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} y_1 & = & x_1 + x_2 \\ y_2 & = & x_1 - x_2 \\ \end{eqnarray}

It should be obvious that the derivatives are simply:

(3)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} y_1' & = & x_1' + x_2' \\ y_2' & = & x_1' - x_2' \\ \end{eqnarray}

Rewriting the equations in terms of y instead of x:

(4)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} y_1' & = & (x_2 - 2x_1) + (x_1 - 2x_2) \\ & = & -(x_1 + x_2) \\ & = & -y_1 \\ y_2' & = & (x_2 - 2x_1) - (x_1 - 2x_2) \\ & = & -3(x_1 - x_2) \\ & = & -3y_2 \\ \end{eqnarray}

I will define the derivative with respect to t, but it could be any variable:

(5)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} y' & = & \frac{dy}{dt} \\ \end{eqnarray}

Solving the equations is quite simple since they are separated:

(6)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} y_1' & = & -y_1 \\ \frac{dy_1}{dt} & = & -y_1 \\ \frac{dy_1}{y_1} & = & -dt \\ \int \frac{dy_1}{y_1} & = & -\int dt \\ ln(y_1) & = & - t + D_1 \\ y_1 & = & C_1 e^{- t} \\ \end{eqnarray}
(7)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} y_2' & = & -3y_2 \\ \frac{dy_2}{dt} & = & -3y_2 \\ \frac{dy_2}{y_2} & = & -3dt \\ \int \frac{dy_2}{y_2} & = & -3\int dt \\ ln(y_2) & = & - 3t + D_2 \\ y_2 & = & C_2 e^{- 3t} \\ \end{eqnarray}

Then we can rewrite the equations in term of x:

(8)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_1 & = & \frac{y_1 + y_2}{2} \\ & = & \frac{C_1 e^{- t} + C_2 e^{-3 t}}{2} \\ & = & B_1 e^{- t} + B_2 e^{-3 t} \\ \end{eqnarray}
(9)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_2 & = & \frac{y_1 - y_2}{2} \\ & = & \frac{C_1 e^{- t} - C_2 e^{-3 t}}{2} \\ & = & B_1 e^{- t} - B_2 e^{-3 t} \\ \end{eqnarray}

This method works if you can create a new separable system of equations.
Another method is sometimes called the elimination method:

(10)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_1' & = & x_2 - 2x_1 \\ x_1' + 2x_1 - x_2 & = & 0 \\ (\frac{d}{dt} + 2)x_1 - x_2 & = & 0 \\ x_2' & = & x_1 - 2x_2 \\ x_2' + 2x_2 - x_1 & = & 0 \\ (\frac{d}{dt} + 2)x_2 - x_1 & = & 0 \\ \end{eqnarray}

Now replace $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \frac{d}{dt}$ with D:

(11)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} (D + 2) x_1 - x_2 & = & 0 \\ (D + 2) x_2 - x_1 & = & 0 \\ \end{eqnarray}

Combine the two equations to eliminate one variable.
To remove $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_2$ multiply the first equation by (D + 2) and adding it to the second:

(12)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} (D + 2)^2 x_1 - (D + 2)x_2 & = & 0 \\ (D + 2) x_2 - x_1 & = & 0 \\ \end{eqnarray}

The result is:

(13)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} (D + 2)^2 x_1 - x_1 & = & 0 \\ (D^2 + 4D + 4) x_1 - x_1 & = & 0 \\ (D^2 + 4D + 3) x_1 & = & 0 \\ \end{eqnarray}

Solving the equation is then trivial:

(14)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} D & = & \frac{-4 \pm \sqrt{4^2-4*3}}{2} \\ D & = & -2 \pm 1 \\ \end{eqnarray}
(15)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_1 & = & C_1 e^{- t} + C_2 e^{-3 t} \\ \end{eqnarray}

By the symmetry of this problem, if you were to solve for $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_2$ you would get:

(16)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} x_2 & = & C_3 e^{- t} + C_4 e^{-3 t} \\ \end{eqnarray}