Question

(a) Show that if yp(t) is a solution of duydu 2 + +qy = g(t), then kyp(t) is a solution of

Answer 1

Sol? @ Since Ypct) is a solution of deporte + palete + ay = g(t) , therefore e n + papie + 2 dp = get) 0 Let Kyplt) = m(t), then (where K is a constant) of lk 3P) = Kette steder og et od ukap) = key Now, multiply equation G by K, we get Karten + KP adresa + Kq jp = kget) det + p dn + qm = kg(t) +2 dt² at

I rzaysa (778035 = R(8 +09+0) 78975 = Rp + Hoy te te gute Hence, ky plt) is solution of het hele pode 2 + 2y = kg (t). (7.78* = + pa je uspony o w€

- Auxilary equation is mat6m +8 =0 (m+4) (m+2) =0 m = -2,-4 Complementary solution (say ye) is Tyc = qe-2t + ca exht I where c, fcq are arbitrary - constants. Particular solution (say Up) is fp = o Core (scoot) = s. meu cate con(t) {up = BCD, cos(at) can be calculating by replacing sa by -x?, provided denominator is non-zero. } 4 - 5 – 5 Cos(t) I coolt) |+60+8 I +6D -17-6D 1 ) cost = cost 5 (7-6D) 49-360² Op = 167+6D)(7-6D)) Jp = 47.62 cost - ABG (1-60) Cost Up = ans ( 7cost-Go cost) = (1 cost - 64-Bint)) Jp = t (7coot + Gaint)

So, general solution (sayy) of given differential equation is y = Yc + Yp y = cie-2t +gecut + (76o8t + 6 sint) where calca are arbitsary constants. Ans.