Question

2. Use variation of parameters to find the general solution y and the particular solution yp. 6) y" + 2y' +y= .73

Answer 1

Sol criven, go + 2y + y = 1 W = e me to be a trial solution of reauced equation 1. Then auxiliary equation is m² + 2m + 1 = 0 2 m+ 1)2 = 0 mar1,-) The complementary function is (4+42) é 2 where Gk le are ansit- sary constant. To find . I (Particular entegrue) Us, y = ex and I = xé a xez -22 then, WCY, 72) = e 22 - 2e-2 tre 22 én-ren - e-22 i I, 92 fo are linean. early independent. Led, the porti cules solution up vi e 2 + ų ren ; where , 22 are the function of x. y = v'er-v, ex + xente ex- exer we take, lo such that, Wertunero I - e, en + q (ex-xen) 2 - - 1 *+ c + -- 7 - ) (2) tre-2 ex 23 from the given equation, y + 2yp+yp = l / 23 A-vier + ₂ Cerexed).- By cross multiplication, re e-2/3 -e-22 e-22 2 - e22 2

Y = -1 22 and r = 11 12 2) U = ana V 11 2 22² (Integrating) Jp = 12. e2 Jaz xez 222 Yp e · 2 22 .. the required general solution of the given equation is ga er? (22) e-nt 22 chure e e ese arbitrary Censternt.