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