# Differential Equation eigenvalue!!

• Jun 9th 2009, 04:53 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:

${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...

Many Thanks....
• Jun 9th 2009, 05:38 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:

${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...

Many Thanks....

How did you arrive at your constants. I obtained

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