+0  
 
0
431
3
avatar+2353 

Define the system of equations;

$$x_{n+1} = x_n - y_n$$

$$y_{n+1} = 2x_n + 4y_n$$

so if we take $$A = \begin{pmatrix}
1 & -1\\
2 & 4
\end{pmatrix}$$

and define

$$\begin{pmatrix}
x_{n+1} \\
y_{n+1} \\
\end{pmatrix} = \begin{pmatrix}
1 & -1 \\
2 & 4 \\
\end{pmatrix}\begin{pmatrix}
x_n \\
y_n \\
\end{pmatrix}$$

Then given that the eigenvalues and a corresponding eigenvector to the eigenvalue can be defined as

$$\lambda_1 = 3$$

$$\mu_1 = \begin{pmatrix}
1 \\
-2 \\
\end{pmatrix}$$

$$\lambda_2 = 2$$

$$\mu_2 = \begin{pmatrix}
1 \\
-1
\end{pmatrix}$$

(I already checked this)

Is it then true that the general solution can be defined as

$$x_n = 3^nc_1 + 2^nc_2$$

$$y_n = -2*3^nc_1 - 2^nc_2$$

Reinout 

reinout-g  Jun 2, 2014

Best Answer 

 #3
avatar+889 
+5

Hi Reinout

Yes, agree with your result, with

c1 = -x0 - y0  and  c2 = 2x0 + y0 .

Bertie  Jun 2, 2014
 #1
avatar+26971 
+5

Perhaps I've misunderstood, but if c1 and c2 are initial values (n=0) for x and y then (for n = 1):

reinout

These don't look the same to me!

Alan  Jun 2, 2014
 #2
avatar+2353 
+5

Hey Alan,

you're right, but they are not initial values.

They are constants which can be calculated for given initial values for x and y

reinout-g  Jun 2, 2014
 #3
avatar+889 
+5
Best Answer

Hi Reinout

Yes, agree with your result, with

c1 = -x0 - y0  and  c2 = 2x0 + y0 .

Bertie  Jun 2, 2014

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