MATH2010 - Analysis of Ordinary Differential Equations Final Examination, Semester 1, 2019 2. (15 marks) Consider the system [y"(t)+3y(t) =(r(t) y(0) = y'(0) = 0, forcing term where Jo t<1, = u(t-1) (,Heaviside 1 f.") r(t) = 1 t ? 1 . It besico (a) Find the Laplace transform of y(t). (b) Compute its inverse Laplace transform to find y(t). 1 7 advanced) Find the transfer function and use the convolution theorem to compute y(t) (note that this is an alternative way to compute the result found in (b) ). a) d(y")= 52 [(s)-sy(0)- y(0) = 52 Y(s) d(r)= {(ult-1)= = === Rb) (2nd shifting thm.) Transform the entire egu .: s2 1/(s) + 3 Y(s) =R(s)= 1-5 [(s) (s2 +3) = R(s) => 1 Y(s) = 52 +3 R(s) = 5(52+3) C 1 -5 = , Q (S ) . algebraic cpu .. for -1 b) goal is to compute y(t) = > 1 -s = s(s2+3) e second shifting them: d (ult-4) {(t-()=F(s)e- Hs u(t-k) { (+-k) =/2-1 (F(s) -(s) 1 =>(+) = == (5(52+3) =)=4(t-1)-1 /5/5 s/s2+3) (7-1) Page 4 of 8 = > need to find -1 2 s(s2+2)
MATH2010 - Analysis of Ordinary Differential Equations Final Examination, Semester 1, 2019 1 (Working Space Only) S(s2+3) = > use partial fractions : Simple factor irreducible Square 1 A = + Bs+ C S (s2+3) 5 S2 + 3 1 . (52+3) s 1 = A(S2+3) + (Bs+C)s 0: 1=3A =A= 1 = (A +B) s2 + Cs + 3A O(s): 0 = C 0(2): 0=A+B => B =- A => 1 5 S2+3 - S (s'+3) = 1 3 5 1 1 1 4 3 52+3 -1 => L 5( ( 1 (+3) =2"(/5- 52+3)=/(1-co)(rt) 11 => >(z) = u (t-1)= (1-cas(23/t=) 1 1 R ( s ) c) I(s) = S2+3 -Q(s). transfer function + convolution thm : frp=gof = (f(T) 9(+) de < (frg) = 2(f). (g)=(F(s) (s) => L' (F(s) (s) = 18 g our goal is to compute y (t ) = {" ( Q (s ) ( s ) = 98+ and r (t) = 4(t-1) 2377 1 1% (t) = 2-1(52+3)= 2 = sin (13)+) 13 y(t)= 2xr = [u(2)/2 sin (23)(+2) $/T = 0 13 s2 +3 t 0 1 next page Page 5 of 8 KLI +>1
MATH2010 - Analysis of Ordinary Differential Equations Final Examination, Semester 1, 2019 (Working Space Only) + (>1: y(+) = Staan (75H-2) dt = 1 + = 12 cas (23(t-C) | =(1-cos(25)(2-1) 11 5 + 1 10 - - Page 5 of 8