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avatar+2354 

Define the system of equations;

xn+1=xnyn

yn+1=2xn+4yn

so if we take A=(1124)

and define

(xn+1yn+1)=(1124)(xnyn)

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

λ1=3

μ1=(12)

λ2=2

μ2=(11)

(I already checked this)

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

xn=3nc1+2nc2

yn=23nc12nc2

Reinout 

 Jun 2, 2014

Best Answer 

 #3
avatar+893 
+5

Hi Reinout

Yes, agree with your result, with

c1 = -x0 - y0  and  c2 = 2x0 + y0 .

 Jun 2, 2014
 #1
avatar+33658 
+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!

 Jun 2, 2014
 #2
avatar+2354 
+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

 Jun 2, 2014
 #3
avatar+893 
+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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