Question

Determine the form of a particular solution for the differential equation. Do not solve. y" - 18y' + 82y = et + tsin 2t - cos 2t The form of a particular solution is yp(t)= (Do not use d D. e. Ei or las arbitrary constants since these letters already have defined meanings.)

Answer 1

Given that the differential equation is y" - 18y'+82y= 26t+t sin (27) - COS(2+) (1) Equation (1) is a 2nd order linear non-homogeneous differential equation. Therefore the general solution of the differektial equation & in y (t) = Y c (t) + Up (t) where ye (t) is the complementary function of the corres ponding homogeneous differential equation of (1) and yo(t is the particular solution of (1). The homogeneous differential equation of 4) is y" -- 18y'+82y=0 Let Yc (t) = emt be any solution of differential equa (2) "Yé = memt mremt since ye(t) is a solution of (2), we have ya -1896 +829€ = 0. m-18 -18m +82) emtzo, (: Ye=emt Ye=ment y elment m 18m+82-0, ( emt+o) This is a quadratic equation in m. So we have 18 & J (18) 2 4x82 → Żm ma 2 2 1821324-328 18+ 34 2 18 +255 2. = gti, izni 1.

.. The complimentary fimetion of (2) is Yelt) = est(acos (t) + G sin (t), G, and q being arbitrary constants. We see that the terms in the right hand side. of the differential equation (1) does not involve in the complimentary function Yc (t) for the root m= 9 ti, so by method of undetermined coefficients wwe take the particulas solution of the form +(B+Ct) (F Sin (2+)+G WS(21)) +(H cos(20+ L sin (2) where A, B, C, F, G, H and L are constants to be determined . ... The form of the particular solution of equa (1) is ypte= (Aest (B+C(F Sin (2+ G cos (ex+ (Hcos (21)+Lsio (10 Yp (t) = Ae6t