Question

Please show how to solve. Correct answer shown.

Use variation of parameters to find a general solution to the differential equation given that the functions y, and y2 are linearly independent solutions to the corresponding homogeneous equation for t>0. - 2t + ty +(2t - 1)x - 2y =ềe -2t, Y1 = 2t - 1, y2 = e - A general solution is y(t) = X X That's incorrect. 1 Correct answer: C1(2t - 1) + c2 e - 2t -12 - 2t 4.

Answer 1

+9"+ (2x-1)7'_29= t ezt - 0 equation 1 can be written y" + (24-1-2y = tēt 2 t t t t are comparing ② with Bt 9"+ petsy't q ty = R it, we have R(t) = telt, ply= (2+ -1, 9H = -2 Now, given that I = 24-1 and = ezt the solution of 6 particular Let J (= Alx) I + Bluz is the station figurat the wronskion of I, and I is given as I h Iyi Solution of o. 27-1 e et 2 -2024 = (2+-1 (-262) - 20 2t = - 4 télt +20t – zot Wa - 4 télt Therefore, Alta Cé. Ce text - It - 4tett & deudt tci AU ē 2t - 8

© BLY= ly, w Est d + 1 tot et-lotet 4 tett - (2-1 dt = 7 { 24 - 4 + ² } Therefore J, H = A (W Y t 8 HZ *&*(2-1) + (-2)(4+24 - tet + 3824 - 12 tett K ty = kelu_futet Hence the general selation of is given as J t = 4h + + 8 4 _ &te a JH = 4 (2+-1) + Get + z et 12 tett or y t = q (24-1) + (2 + 1 Je 4 - 4 t e 24 y Lb= q (21- U+ Gel het het ezt is the general solution of O. But qt & =(3, weget