# ODE1.1

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Solve the system of ODE's: $\begin{cases} x_t = 6x-3y\\y_t = 2x+y \end{cases}\,$

Guess the solutions

$x = Ae^{\lambda t}, x_t = \lambda A e^{\lambda t}\,$
$y = Be^{\lambda t}, y_t = \lambda B e^{\lambda t}\,$

Plug these guesses into the original equation.

$\lambda A e^{\lambda t} = 6 A e^{\lambda t} - 3 B e^{\lambda t}\,$
$\lambda B e^{\lambda t} = 2 A e^{\lambda t} + B e^{\lambda t}\,$

Write this system of equations in matrix form.

$\begin{bmatrix} 6-\lambda & -3 \\ 2 & 1-\lambda \\ \end{bmatrix}$$\begin{bmatrix} A \\ B \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\,$

Find lambda by taking the determinant of the matrix.

$(6-\lambda)(1-\lambda) + 6=0\,$
$6-7\lambda+\lambda^2+6=0\,$
$\lambda^2-7\lambda+12=0\,$
$\lambda = \frac{7 \pm \sqrt{49-48}}{2} = \frac{7 \pm 1}{2} = 3,4\,$

There are two values for lambda, so there will be two solutions and the sum of these two solutions is the general solution.

If $\lambda = 3\,$,

$3A-3B=0\,$
$2A-2B=0\,$

Add these equations to get
$A-B=0\,$, so $A=B=C_1\,$

One solution is $x=C_1 e^{3t}, y=C_1 e^{3t}\,$

If $\lambda = 4\,$,

$2A-3B=0\,$
$2A-3B=0\,$

These equations imply
$B=(2/3)A\,$, so if $B=C_2, A=(3/2)C_2\,$

Then the general solution is

$x = C_1 e^{3t} + (3/2) C_2 e^{4t}\,$
$y = C_1 e^{3t} + C_2 e^{4t}\,$

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