Math Help Forum - View Single Post

Danny · #2 June 9th, 2009, 06:38 AM

Quote:

Originally Posted by althaemenes

Hi, I am having trouble with the following problem. I dont know where I am doing it wrong... please help me out...

Problem:

Apply the eigenvalue method to find the particular solution to the system of differential equations

which satifies the initial conditions

___________________

Attempt:

${A} = \begin{pmatrix}{6}&{7}\\ {5}&{8}\end{pmatrix}\\=> \left|A-\lambda I \right| = \begin{pmatrix}{6-\lambda}&{7}\\ {5}&{8-\lambda}\end{pmatrix} = 0\\=> \lambda = 13 \quad{or} \lambda = 1 \\$

So I found the eigenvalues and solved for eigenvector.

So for lambda = 13 eigenvector $\begin{pmatrix}{1}\\ {1}\end{pmatrix}$

for lambda = 1 eigenvector $\begin{pmatrix}{1}\\ {-5/7}\end{pmatrix}$

$x(t) = {C_1}\begin{pmatrix}{1}\\ {1}\end{pmatrix}e^{13t}+{C_2}\begin{pmatrix}{1}\\ {-5/7}\end{pmatrix}e^{1t}$

Now for the given initial conditions I finally got:

x1 = -16/3*e^(13t) - 7/3*e^(t)
x2 = -16/3*e^(13t) + 5/3*e^(t)

But its wrong...

Please help...

Many Thanks....

How did you arrive at your constants. I obtained

$x = - \frac{2}{3}\begin{pmatrix}{1}\\ {1}\end{pmatrix}e^{13t}- \frac{7}{3}\begin{pmatrix}{1}\\ {-5/7}\end{pmatrix}e^{t} = - \frac{2}{3}\begin{pmatrix}{1}\\ {1}\end{pmatrix}e^{13t}- \frac{1}{3}\begin{pmatrix}{7}\\ {-5}\end{pmatrix}e^{t}$