Question

Diff. Eqn variation of parameter method

(4) you so Solve the equation by method of variation of Parameter?

Answer 1

Solution : y(A) = 5x ܟܙ To find C.F. The auxiliary equation is mt zo CY, m=0,0,00 The Complementary function is + eza²+ Ca&3 C+C₂2t To find P.I. where by Replace ci by Vilay , ca by U₂(2), C3 by V3(2), Ca by 4 (2) and take afp as a particular solution Jp = U194.1 + Vzcay.X + V3 (4.170(23.23 Now, vica, Ulay, Vicky and va lay are found by Solving: WX8.2 + uzcay.x+14.27+ Vacay.x}=0-> V'lay. 0 + ( a1 + Uz'(. 2 + V (4.34-0 21+ 3 V/ 10.0 + Vez' (04.0+ vzran.2+124623.6220 LO vilcoy, ot uzca.O + V3(34.0 +V4(04.6=54 Fromo, V4'lay 동지 Hence, valas / Enda = 52 52² H

Now, from Vz(29.2 + x).6x20 [:V46 Ex] on, 2 Ugray s-582 ar Uslay = -5x2 Hence, Uzay s Sesta ---- 523 6 From ③ > Vz'(a1 + (-3%)-2x +(Ex).32+=0 or, U2(M = 5x² - 523 or U₂ (2) 5x3 = Hence, 1269 x3) da 24 524 - F. 24 = 8 From 2, W/(24.1+ (x3).x+6x9.2+(2x).x=0 m. vyras + Ime - {%$+$240 - sx4 SE5x) dx = F xs on Yaxy Hence, y cay = an4x5

Accordingly, Ypa - Insassa + 5 - 40 41585-2005 tlove 24 2575-2475 25 24. 24 ... The con complete primitive is y = G+qatga + G234 25 24