Question

Use the method of variation of parameters to find a particular solution of the differential equation

y′′−11y′+28y=162e^t.

Answer 1

we have Y"_lly' +984 =162 e It can be written as, (D2_11D+98) y = 169 et where D = 24 and R=16get It's Aurilliary equation is, ma_11 m+ 98=0 » m²-am-7m+98 =0 27 m= 4,4 ma l Hence, complementary function is t . . Giet + Cg ett ☺ Now the Particular solution of ④ is, 1. yp= uy,+ vy, - ② where, y,= eht; Yg = ett be two linearly independent solution of homogenaus differential equation yol -lly' +98720. Woon skein of y, and Y₂ be, Now, W (91, 92) 19 Yl I ett - ett llt Yeux 7e =

Now, dt -3t. C -32 dt Se --suse tut sie anova SR 162 people at and _ dt sy sige omst 3 e" v = -9 ést put u and r we get, Particular solution as, . úy, tvy, 1. = 18 e 3t eut qeft ezt - 18et get 3p = 9 et I. Hence the resuld. I NOTE:- o nim, Also, complete solution is, 4cta a ent tczett+qet). utt.