Question

Find a general solution to the differential equation using the method of variation of parameters. y"' + 4y = 3 csc 22t The general solution is y(t) =

Answer 1

of putting R=0 + Method vacation of paumeter D. E dy & Pay & Qy ER, P, Q, R are funct oft dt at find C F by putting say CF = Cigi + C2Y 2 w ly, Celalte wronskein y2 Y Y 2 h • Dolcetion is ? y(t) = u(+) y 1 + Uctly2 where ult)=-119 ₂ R dt tel y, R dt & c" W uC+) = je

+ 4y = 3 cosec at Now given question is y to find CF, let y " + 4y =0 l-e m² + 4 =0 m² = - 4 m= IPA CF = G conat+ C sin 2t. i-e y, =conat and Y 2 = sin 2t. Conat sinat 1-281nat geopet 2 con ²24 + 2 sin 224 = 2(con224 + sin22t) = Now to find our solution, we need to calculete VC*) Lių(*) ut and Ye R. d t + c ut - at tch sinat . comenat at + c 2 - 2 s sind sin²t dt to sinat copec (2+) at tc uring Scopee dedx=- 2 = I scorer - log/copecret cotel Iconec (+) +Co+(2+)) we have ult) = 1 4 te!

and now to find act) UC- Y, R at + c = 1 f con (2+) conec? (2+) at to" con (2+). sin²(24) = at & c 2 det sin (24) = 4 2008 (24) dt = du Con (24) at = du 2

we hone H du + c 2 २५ tel 4 ru creplace u by sin(at) I sin lat 4 7+" Lect) = ta". 489(2+) y (+) = Noth the sojection will be given by u(t) J + v (A) Y Z Gü Jog.conce(217+ co+(24) 1) a condt. sin at + c'coort tc" sinat = + 4 sinat