Question

To show that (0,0) is always an unstable critical point of the linear system:
x'=μx+y
y'= -x+y
where is μ a real constant and μ≠1

To find where (0,0)is an unstable saddle point.

To find where (0,0)is an unstable spiral point.

Answer 1

looks like this is different Q

Answer 2

yes

Answer 3

memory is a bit fuzzy ... hold on this takes a time.

Answer 4

the matrix is
u 1
-1 1

Answer 5

get the eigen values from here

Answer 6

That is the matrix which I found. (u+1)(u-3)>0 and u+1<0 If d=determinant and t=trace of matrix then t^2-4d >0 (discriminant of quadratic) and d<0 are the criteria to classify the critical points. Can you give the values for the saddle point and also for the spiral point now?

Answer 7

http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iv-first-order-systems/qualitative-behavior-phase-portraits/

Answer 8

http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iv-first-order-systems/qualitative-behavior-phase-portraits/LinPhasePorCursor.html

Answer 9

for unstable spiral .. you need imaginary part ... with positive real part

Answer 10

since this is linear system ... all critical point will be centered in 0,0

Answer 11

for unstable saddle ... you will have ... one +ve part and one -ve part.

Answer 12

you will have no imaginary part.

Answer 13

kinda remembered ... i skipped in the middle of non linear system