Question

Q1. Below is the bifurcation diagram for a first-order differential equation. y- 2 k=0 k-4 (a) Write down a first-order differential equation that would have this bifur- cation diagram. tative behaviour of solutions occurs. lution if (b) Find bifurcation values of k, i.e., values of k where a change in the quali- uit so.

Answer 1

y=妬 0 4 Clearly, For there ezists only one equilibrium point, .e, y 2. Likewise, at kthere exists two eqilbrium points, y 2 and y Ξ0 With 0 as repeated equilibriun point since it's on parabola y k, k0, And, For k there exists three equilibrium points y, y-VE and y VESo, the ode is givena du dt (b) Clearly, FYom theucation diagyam, the volues of k whem change in the qualitative behaviour of solution CCs s at 0 and 4 At k-o the number of solutions increases from 1 to 2 as k is increased from ne gohive to positive (efer bipuration diagram) . And, at k: 4 bheve the number of soluions decreases Fiom 3 to 2 also the nature of the behoiour changes at equilibri um point y 2, Xom sink to saddle

(C) ce, from port Ca) the First crder difperntial equation IS dt (i) Substitute k:-2, gives dt Clearly, From the bipumation dogran For k<o there s only one equilibrium point, i.e., y: 2 and since the divechon of arrous points towords it therefore, it is a sink. Hence, the equilibrium point y 2 is a Sink or 2 (ii) Subs tituhe k2, gives dt Clearly, From the bifucation diagyam FOY ock<4 there ezkts three equilibrium points, ie., y--5/2,2. For 2 divection of aTDAs points towards it, therefox, it is a sink, For yi direchon of anws points away From it, the efoe, it is a souce, and for y 2 direction arrows points towords it,the vepone, it is a sink (iii) Subsh'tuteks, gives

du dt Clearly for biguration diagram for k4 there exists thre equilibrium points, y--/5, 2,5. For y--B is a sink, y=2 is a source and y=/5 a sink.