# Homework Help: Diff Eq linear systems help

1. Dec 15, 2008

### 2slowtogofast

i am trying to study for a DE test and was wondering if you could help with some problems

dY = -2 1
dt 0 2

i found the eigen values to be lamda = 2

i need some help on how to find the eigen vectors is this a reapeted root case and if it is how do i start off

would i go AV = (lamda)V then solve form there and in this case there would only be one eigen vector. I aplolgize on my lack of knowledge on typing in math symbols.

2. Dec 15, 2008

### Dick

The eigenvalues aren't both 2. There are two linearly independent eigenvectors. Try that again.

3. Dec 15, 2008

### 2slowtogofast

i will give it another go

4. Dec 15, 2008

### 2slowtogofast

if i subtract the identiy matrix multiplied by lamda from

-2 1
0 2

i get the new matrix

( to make this simpler im going to use L for lamda)

( -2 - L ) 1
0 (2 - L)

then if i take the determinate of this matrix and set it equal to zero i get

(-2 - L)(2 - L) - 1(0) = 0

L^2 - 4 = 0

L^2 = 4

L = 2

i must be making a mistake somewhere can some point out where thanks for the help

5. Dec 15, 2008

### Dick

Stop with (L-2)*(-L-2)=0. It's already factored. The roots are +2 and -2. Likewise for L^2=4.

6. Dec 15, 2008

### 2slowtogofast

ok thank you so then +2 and -2 are my eigen values thank you for the help

7. Dec 15, 2008

### 2slowtogofast

using those values i calculated the following eigenvectors.

AV = LV

from this i get the eqautions

-2x + y = 2x and 2y = 2y then i get

y = 4x and y= y

if i let x = 1 i get

V = 1
4

for my first eigen vector

then for the second i get

AV = LV then i get the eqns

-2x + y = -2x solving i get y = 0

and the other eqn is 2y = -2y im a little confused on how i can get my eigen vector from these two eqns. y = 0 but how can i get the x val

8. Dec 15, 2008

### Dick

You've got that y=0 and x=anything. There's no equation for x. Pick an 'anything'. like x=1. That's works doesn't it?

9. Dec 15, 2008

### 2slowtogofast

ok i was kinda thinking that but was a little unclear but thank you. the next part i am working on is for each eigen value pick an associated eigen vector V and determine the solution Y(t) with Y(0) = V

could you please help me start this one off im confused on that an what it meanse to pick an eigen vector cant i just use the ones from the prior part.

10. Dec 15, 2008

### Dick

Ok, pick (1,4) (I'll write the vectors that way instead of in column form). Assume the solution has the form (1,4)*w(t) (where w(t) is an ordinary function, not a vector). Substitute that into the DE. Can you solve for w(t)?