Question

4. Consider the following autonomous equation: 3/4 (4-2 -3)0 W Con Choose any value for t. ay = (4 - 1)(y - 3)2 (a) (1 points) Find the equilibrium points ie (,2), (+,3) PER are equilibrium points .: Y= 2 43 (b) (4 points) Draw the phase line + 112 341-1-2-3 12-11- 2 0141214-21-3 Y 1200-121200562 34 20 20 YOOOOOOOOOOOOO -4 (c) (2 points) Classify the equilibrium points as stable, semistable, or unstable. The given equilibrium points are semistable (d) (3 points) Suppose y(t) is a solution to the initial value problem I, B) = (4 - v)(y - 3) y(0) = Find the limit lime-te y(t). Justify your answer. liny (1) lim ( y) (4 7 -- X As y Increases, it becomes y² y ² boyu

Answer 1

Guren autonomous equation The equilibrum points are l.e. when dy becomes zero yh) (4-3) = 0 & (4.72) =0 on (4-3) = 2 2) Yu=4 y=3,3. o y = 250 k ron ... The equilibrium points are 252 and 3 of multiplicity 2.

6-2, 2 & 3 are equilibrium points, so there will be no change of solutions. when y4-2, dy a -22Y<2 oso 22 y 23, d4 0 973, d7 Lo

Then the phase line is given by 39 dy ko A 40 -2 where anrows on aedicate solution is moving in which direction ! (C) At y=3 &, therefore this is unstable. At ya 2, 2, therefore it is stable: At y=-2, 1-2, therefore it is unstable. Br?

d) autonomous differential Given is dx = (449-3) It is vany hard to find yats from solving this diffenential equation. So, we have to find the behavious of y(t) as a> a aceonding to the initial conditions given Here y(a)=Y, which lies in (0,2). Herre, already seen 2 is the stable equlibrum point. So, when t o , if y(0) =Ys, then this y H) surely. approaches to the value 2. Hence lim y) = 2, when yo=16. ta & also (4-225) (- 3 0.