1. For the following find the steady states, and then...

1. For the following find the steady states, and then determine whether they are asymptotically stable, unstable, or indeterminate. Any parameters appearing in the equations should be assumed to be positive. a) {x' = 1 - 2x - y - xy, y' = 3xy - y} g) {u' = e^u - v, v' = uv} b) {x' = y - x^2, y' = y + x^3} h) {u' = au - buv, v' = -cv + duv} c) {u' = 1 + v, v' = u + v^3} i) {S' = 2S - S^2 - (2SP)/(1+S), P' = (2SP)/(1+S) - P} d) {u' = 4 - uv^2, v' = -v + uv^2} j) {S' = -1/2 IS + 1 - I - S, I' = 1/2 IS - I} e) {u' = u^2 - v, v' = 2u - 3v} k) {u' = v - u, v' = (2 - u - v)(1 + v^2)} f) {u' = u^2 + v^2, 2v' = sin(u)} l) {S' = -aES + b(E_0 - E), E' = -aES + (c + b)(E_0 - E)}

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00:01 Okay, so first see the given system is, given system is u equals to 1 plus v equals to f unv and the v and the v that equals to u plus v3 power equals to z unv so now for fixed point points making bracket u and a v goes to 0 and g u navv equals to 0 we get 1 plus v equals to 0 then v equals to minus 1 bracket close so u plus v 3 power equals to 0 u equals to minus v 3 power equals to 0 equals to minus inside a bracket minus 1 whole bracket q equals to 1.

01:40 Q means 2 mean is 3 power.

01:44 Then 1 minus 1 is fixed point.

01:59 Now j inside u and v goes to bracket f, u, v, g, g, g v goes to inside a bracket 0 1 1 3 v square so j after putting the values 1 minus 1 equals to 0 1 1 3 equals to a now a minus lambda i equals to 0 so 0 minus lambda 1 1 3 minus lambda is to 0 so lambda lambda and them minus 3 bracket close minus 1 equals to 0 lambda square minus 3 lambda minus 1 equals to 0 so lambda equals to plus minus root 9 plus 4 upon 2 equals to 3 plus minus root 11 upon 2 so lambda 1 equals to root 13 plus 3 2 and that 2 equals to minus root 13 plus 3 upon 2 so since the point 1 minus and minus 1 minus 1 minus 1 minus 1 point and stable and that 2 is less than a 0 less than a lambda 1 e the given system is u equals to u x square minus v equals to f unvv 2 u is 3v equals to g to u n a v now 4 fixed point making f u and v equals to 0 and g u and v equals to 0 so we get u minus v equals to 0 and v equals u square minus 3 v equals to 0 it will be 2 u minus 3 u squared equals to 0 u inside the bracket 2 minus 3 u bracket 0 so u equals to 0 2 by 3 if u equals to 0 then v equals to 0 and if u equals to 2 3 then v equals to 4 by 9 so 0 into 0 and 2 by 3 and 4 by 9 are fixed point now j uv equals to bracket f u f v g u gv equals to bracket f u f v g u gv equals equal to bracket is 2 u minus 1 2 minus 3 so j after putting values 0 .0 so i had a bracket 0 minus 1 2 minus 3 b now b minus lambda i was to 0 so minus lambda minus 1 2 minus 2 minus 0 lambda plus 3 plus 2 equals to 0 it will be lambda as square plus 3 lambda plus 2 equals to 0 it will be lambda plus 1 bracket lambda plus 2 equals to 0 lambda equals to minus 1 bracket minus 2 bracket so so 0 and a 0 is a stable, table node.

09:14 So we can see lambda 1, under 2...

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