Question

differential equations matrice problem

The matrix [−43−982148] has eigenvalues λ1=−1 and λ2=6. Find eigenvectors corresponding to these eigenvalues.
v⃗ 1=
⎡⎣⎢⎢ ⎤⎦⎥⎥
and v⃗ 2=
⎡⎣⎢⎢ ⎤⎦⎥⎥


Find the solution to the linear system of differential equations {x′y′==−43x+21y−98x+48y satisfying the initial conditions x(0)=−6 and y(0)=−15.

x(t)=
y(t)=

http://i.imgur.com/f3HZi.jpg

Answer 1

So where are you stuck?

Answer 2

second one

Answer 3

i couldn't achieve a true solution,webwork doesn't accept it

Answer 4

why are you staring like that Turing ?:P

Answer 5

watasiva kira

Answer 6

nice to meet you

Answer 7

Keep the post relevent

Answer 8

I wouldn't say irrelevant, but can you clarify it with either Latex or the equation feature ruhmshalle ?

Answer 9

guys,i want to learn how to solve this problem.

Answer 10

can you help me instead of having a conversation please

Answer 11

you are free to express your opinion in the chat

Answer 12

@ruhmeshalle
Like I said above, I cannot read it. Can you use Latex or the equation feature?

Answer 13

http://i.imgur.com/f3HZi.jpg

Answer 14

it should be ok now

Answer 15

I would try this, but I am in the middle of reviewing ODE's and haven't done systems in a while....

Answer 16

thank you tough

Answer 17

though*

Answer 18

6 views but no reply, sad...

Answer 19

JamesJ knows I'm sure, but he is biding his time...

Answer 20

Try solving it using D-operator

Answer 21

(D+43)[x] -21 y =0
98 x +(D-48)y=0

Answer 22

sorry, I stepped away.

Answer 23

Ok. So you've found the eigenvectors v1 and v2 corresponding to L1 and L2 (L for lambda)?

Answer 24

where L1 = -1, L2 = 6

Answer 25

If so, then write X(t) for the column vector ( x(t) y(t) )^t, where that last t is transpose.

Then the general solution of the system is

X(t) = C1.v1.e^-t + C2.v2.e^6t

where C1 and C2 are arbitrary constants.

Answer 26

Following so far?

Answer 27

yes

Answer 28

i'm back

Answer 29

Ok. Now, what you need to do is find C1 and C2. The way to do that is to use the initial conditions, x(0) = -6 and y(0) = -15

Answer 30

Substitute those into the general form. I.e.,
X(0) = ( x(0) y(0) )^t = ( -6 -15 )^t
then
X(0) = C1.v1.e^0 + C2.v2.e^0
= C1.v1 + C2.v2

Answer 31

Hence you have a system of two linear with two linear variables. Solve for C1 and C2.

Answer 32

thank you, james. it really helped, i will try it now.

Answer 33

If you're still a bit shaky on the theory here, you might find this lecture and/or the ones immediately around it helpful:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-25-homogeneous-linear-systems-with-constant-coefficients/

Answer 34

Are you still OK with the previous post:
http://openstudy.com/#/updates/4ef5048ce4b01ad20b507845
which has since been deleted.

"The vectors v? 1=[-32] and v? 2=[1-1] are eigenvectors of the matrix [-6 8 -1214]. Multiply them by the matrix to find their eigenvalues. ?1= ?2= Use your answer from part (a) to find the solution to the linear system {x'y'==-6x-12y8x+14y satisfying the initial conditions x(0)=-5 and y(0)=3.
"
It would be done in exactly the same way James did earlier.