Question

322-M2 4) Find the inverse - transform of F(z) = (2-1)(2-2M)' (15 marks) 0 t<-M/2 M <t< - 5) Show that the Fourier transform of function f(t) sin 7 s (10 marks) au 6) Show that u = ln(x2 + xy + y2) satisfies the partial differential equation x x ди +y 2. (7 marks) au 7) Solve the partial differential equation = e-cos(x) where at du x = 0, at =tet ax at and t = 0, u = sin(x). (13 marks)

where M=7

Answer 1

Sol A Given fonction is f(2)= 32²-MZ (2-1) (2_201) where M = 7. F(7) 32²_77 (2-1/2-11) Sol F(A) = 32- (2-012-14) Z B + F(2) A (2-1) (7-14) (say) 3Z- (7-1)(2-14) A (2-14) +B(2-0) (2-1) (214) ar, 32-7= A (2-14) + B(2-1). 4 Putting 2=1, AL-13) = 3-7 or, A= 1 / 3 and putling = 19, B (13) = 42-7. ar, B= 35 13 35 So, F(2) + 13 (2-14) 7 13 (2-1)

Then its becomes /2 e-istat. sh 7{f(t)}= F(S) = √27 Ah e-ist p2 e Tat لا si -72 6- 15/2 This -is TET - is e 7/2is é-ist is 2 : D is 2T this & this]. S e 2 [ 21 F=M, so s (1 1 / 2 sinths 1.2.FC) - Esin (45) whore sak is Fourier constant s Am 6 - su - som Here u= ln (x²+ xy + y 2) (2x+y) [ Poctial derivative x2+xy + y2 warto x sy and su (ay+x) ry x2+xyty 2 + as x du 2x su 29

X, (2x+y) x2+ y2 + xy + y (2y+ x) x2+ y2 + xy = 2. 2 (x2+xy + y 2) (+2+ xy + y2) x du y au Proved. - 2. + 22 ay @sol? Given that azu e-t axit cose, where x=0, 34 = tet, and to le-sind, gru Integrating with respect to reweget. anat are It Jetcosxdatel Ju e-t sin & tci ri st when Da Now x=0 dt = tet Ju e-t sind &tet. at tet - a C1 C2 are integraling constant Integrating worto cl=C(+) C2=(26x) u= – et sinx + / tet tea, t we get-

or u=- et sina ttet et tea using u=sing weget t= 0, sink = – Sina - 1 +(2 or C2 = 2 sinatl. u = et sin se tet(t-1) + zsinat! sinx (2-et) tet (t-1) +1. ar, u Aus