homogeneous differential equations

[logistic_guy]

here is the question

Solve $\displaystyle \bold{X}' = \begin{bmatrix}3 & -18 \\2 & -9 \end{bmatrix}\bold{X}$.

my attemb
$\displaystyle \bold{Z} = \bold{A} - \lambda\bold{I} = \begin{bmatrix}3 & -18 \\2 & -9 \end{bmatrix} - \lambda\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} = \begin{bmatrix}3 & -18 \\2 & -9 \end{bmatrix} + \begin{bmatrix}-\lambda & 0 \\0 & -\lambda \end{bmatrix} = \begin{bmatrix}3 - \lambda & -18 \\2 & -9 - \lambda \end{bmatrix}$

det$\displaystyle (\bold{Z}) = (3 - \lambda)(-9 - \lambda) - -18(2) = -27 -3\lambda + 9\lambda + \lambda^2 + 36 = \lambda^2 + 6\lambda + 9 = (\lambda + 3)(\lambda + 3) = (\lambda + 3)^2 = 0$

this mean $\displaystyle \lambda_1 = \lambda_2 = -3$

$\displaystyle (\bold{A} - \lambda_1\bold{I})\bold{K} = \begin{bmatrix}3 - \lambda_1 & -18 \\2 & -9 - \lambda_1 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}3 - -3 & -18 \\2 & -9 - -3 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}6 & -18 \\2 & -6 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}0 \\0 \end{bmatrix}$

i can divide by $\displaystyle 2$ to simplify

$\displaystyle \begin{bmatrix}6 & -18 \\2 & -6 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}3 & -9 \\1 & -3 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}0 \\0 \end{bmatrix}$

i'll use the equation $\displaystyle k_1 - 3k_2 = 0$ from the matrix

$\displaystyle k_1 = 3k_2$

i'll chose $\displaystyle k_2 = 1$ then $\displaystyle k_1 = 3$

$\displaystyle \bold{K} = \bold{K_1} = \bold{K_2} = \begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}3 \\1 \end{bmatrix}$

that's not goodbecause $\displaystyle \bold{X} = \bold{X_1} = \bold{X_2} = \bold{K_1}e^{\lambda_1 t} = \bold{K_2}e^{\lambda_2 t} = \begin{bmatrix}3 \\1 \end{bmatrix}e^{-3t}$

that only give me one solution. i need a different solution for $\displaystyle \bold{X_2}$ but the problem is $\displaystyle \bold{X_2} = \bold{X_1}$
what shoud i do?

 

[topsquark]

here is the question

Solve $\displaystyle \bold{X}' = \begin{bmatrix}3 & -18 \\2 & -9 \end{bmatrix}\bold{X}$.

my attemb
$\displaystyle \bold{Z} = \bold{A} - \lambda\bold{I} = \begin{bmatrix}3 & -18 \\2 & -9 \end{bmatrix} - \lambda\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} = \begin{bmatrix}3 & -18 \\2 & -9 \end{bmatrix} + \begin{bmatrix}-\lambda & 0 \\0 & -\lambda \end{bmatrix} = \begin{bmatrix}3 - \lambda & -18 \\2 & -9 - \lambda \end{bmatrix}$

det$\displaystyle (\bold{Z}) = (3 - \lambda)(-9 - \lambda) - -18(2) = -27 -3\lambda + 9\lambda + \lambda^2 + 36 = \lambda^2 + 6\lambda + 9 = (\lambda + 3)(\lambda + 3) = (\lambda + 3)^2 = 0$

this mean $\displaystyle \lambda_1 = \lambda_2 = -3$

$\displaystyle (\bold{A} - \lambda_1\bold{I})\bold{K} = \begin{bmatrix}3 - \lambda_1 & -18 \\2 & -9 - \lambda_1 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}3 - -3 & -18 \\2 & -9 - -3 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}6 & -18 \\2 & -6 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}0 \\0 \end{bmatrix}$

i can divide by $\displaystyle 2$ to simplify

$\displaystyle \begin{bmatrix}6 & -18 \\2 & -6 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}3 & -9 \\1 & -3 \end{bmatrix}\begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}0 \\0 \end{bmatrix}$

i'll use the equation $\displaystyle k_1 - 3k_2 = 0$ from the matrix

$\displaystyle k_1 = 3k_2$

i'll chose $\displaystyle k_2 = 1$ then $\displaystyle k_1 = 3$

$\displaystyle \bold{K} = \bold{K_1} = \bold{K_2} = \begin{bmatrix}k_1 \\k_2 \end{bmatrix} = \begin{bmatrix}3 \\1 \end{bmatrix}$

that's not goodbecause $\displaystyle \bold{X} = \bold{X_1} = \bold{X_2} = \bold{K_1}e^{\lambda_1 t} = \bold{K_2}e^{\lambda_2 t} = \begin{bmatrix}3 \\1 \end{bmatrix}e^{-3t}$

that only give me one solution. i need a different solution for $\displaystyle \bold{X_2}$ but the problem is $\displaystyle \bold{X_2} = \bold{X_1}$
what shoud i do?

Check your textbook as to what to do when you have repeated roots.

-Dan