Question

6, (9 marks) For each of the following direction field plots, write down a function f(t. y) such that the differential equation dy = f(t, y) dt could have this direction field. equation could have the direction field shown. In each case, give reasons why you think your differential 3.5 1IITT1 1T1 111I 1 tIFI I 1 111 1111 111I1 11TTII rrrr 2.5 ttttt 1 ! 2 15 Y 0.5 -111 1 -0.5. 1 t -1 1 111-1 tT t 1 11 1 111t 1TTITrtII- -1.5 T1t 1 1 1 1 1 -2 3 10 1 -11 1 -2 7

Answer 1

Given, In the given figure the sign of the direction field changes when we cross the line, y=215 and y=o. So y=0 and y= 2.5 are two equilibrium point of the system (1). So f (try) = ky(y-2.5) since above the line y=0 the slope is ve implies kyo. i f(ty) =ky (3-2.5) with Kyo.

Giren, de f(ty) In the given figure the sign of the slope value changes above and below the line y=0 and ytt=o. So every point on the line y=0 and ytt =0 are equilibrium point. so f (try will be of the formi 3 (19) - K9 ( + ) The value of slope has are sign above y=0 so kko. ../ (+14) = kg (ytt) with KLO,

Given, dy = {(try) In the given figure the sign of the direction field changes when we cross t=0 and t=2. Therefore to and t= 2 are two equilibrium point of the system (1). Also the giren figure remains constant for every valce of y. 7 / 4y) is independent of y. Hence For f (tay) - kt (+-2) tro che tal the slope is positive. 7 klo in f(y) = xt (t-2) with kko.