MATH2010 - Analysis of Ordinary Differential Equations Final Examination, Semester 1, 2011 1. (a) (16 marks) Setting y1 = y, y2 = y' express the second order ODE y" - 4y' - 5y = 0. as a system of DEs y = Ay, y = ( 1) y2 for a suitable 2 x 2 matrix A. Sketch the trajectories of the system in the phase plane, indicating the direction of flow, and classify the type and stability of the critical point at the origin. so have y , ' = y ' = y2 y2 = y " = 4y + 5 y = 5y + 442 = ) y ' = A y . A = (54). 1 (2 -4) -5 : 14-XI = 1-15 42-× 1 = (1-5) (i++) : > = - 1,5 are evalue => crit. PA. at origin is a saddle pt. which is unstable. For st. line trajectoires need the e-vectors: ? =- 1 : _ = (A-XI)= = (5 5)(w) ? u+rzu = (-1) is envector so y2 = = y, . . determinis st. umè trajectory >= 5: 2 = (A-XI) == = (51) () => v=Ju (arrow tourand determinis . (5) in correcto so y2=5y, st. live trajectory ( arrows away from origin ) . - Page 1 of 8
Moreover dyr = yv/y . 0, y == 0 , = 59,+4/2 = ) 0, 92 = - 5/4 y, dy, 1 · I2 4, y; = 0 giving the phase portrait shown y = 5g, 1 VI K y= - y
--- .. MATH2010 - Analysis of Ordinary Differential Equations Final Examination, Semester 1, 2011 1. (b) (16 marks) Find all the critical points for the following nonlinear system. (% ) = ( y1 + y2 - y'} ). not lying on the Then use linearisation to find the type and stability of the critical points which lie in the first quadrant. y, or y 2 axes. ) - y2 3-y, and crit. pts are 81 + 2 - 71 2 =0 } IND . Equ . ? y2 (3-71) =0 . 72 =0 or 1 = 3 given by Jacobian is Lly, y2) = - 1-2y , 1 3 y2- did= = 0 ) Subst. lite ist equ. gives: ¥220: y=y = y,=02 1 11=3 : 12-712y= 6 :. 3 crit. pts ( 8 ) , (6 ), ( ? ) - all lie in Ist. quadrant .. crit. pt. (0): L = L /0,0) = (0) Improper Node - unstable . 1 3 ) a) e-values >=1,3 . crit. pt. (!): L= L (1,0) = F 1 1 0 2 e-value > =- 1, 2 - unstable saddle point. UNS , -5 1 ) Crit. pt. (3) : L = L (3, 6 ) = ( 2) | L-XI / = -5-2 .1 -6 0 = (x+2) (x+3) stable improper Node . .: 02 P = t (A) = - 5 <0, q = det (L) = 670