Question

2. Solve following questions (Inverse operator method) (10) y” + 2y+5y = xsin 2x

Answer 1

Jolution: Given equation is y' +2y' + 5y = x sin2x Let us considey y" + 2y' + 5y = xsin as (D+20+5) y = x 510 2x Auxilary equation is as follows me + 2m +5 Roots of a quadratic equation - M : abtJ6² 4ac Ja ME -24 14- 46015) -=-24V="6 2 M = -2+ M = -(tai If yoofs of characteristic equation are in tui form then general solution Y(e) = crede cos[ue]+e; czem sin(ME) Now, complementary solution is 4c8) = giet cos(at)+ cq et sin (2t ). now particular solution you (sin(2x)) 044 2045

Now put 02-22 you non x sinex P -4+20+5 Se * sin 2x Rationalize the factor with "2D-1" 2D-1 x sin 2x 40²,2 Yo = 20-1. x sin 2xl- 41-4) -1 " Sp = 2074-X Sin 2x y = 20 X sin 21 - X sin 2x -17 D(X sin 2x)= sin 2x +11 (05 2x sin 2x + 2X COS 2X -sin 2x -17 - 9 sin(2x)+ 4X COS(21) – x sin 2x -17 General Solution y = 4,16) + 4p(t): y = G, ecx cos (2x)+cą e-* sin(2x) – +- (2 sin (2x)+ 4x cos(2x)] -X Sin(2x)