# Differential Equation eigenvalue!!

• Jun 9th 2009, 03:53 AM
althaemenes
Differential Equation eigenvalue!!
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

https://instruct.math.lsa.umich.edu/...f2f7c67ce1.png

Attempt:

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

So I found the eigenvalues and solved for eigenvector.

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

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

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

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...

Many Thanks....
• Jun 9th 2009, 04:38 AM
Jester
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

https://instruct.math.lsa.umich.edu/...f2f7c67ce1.png

Attempt:

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

So I found the eigenvalues and solved for eigenvector.

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

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

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

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...

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