Question

[10] 6. Find the general solutions to the equations (a) 2x+y" + 5xy' + y = 0, (6) Y" – 2y + 5y = 3, YIV – 2k²y" + k*y = 0, (c) where k is a positive parameter.

Answer 1

solution aj Given differential equr is 2x²y" + 5*4l +y=0 which is -(1) a homogeneous linear D.E with variable coefficient then a 2dzy ED(P-1) the given equation becomes xdy ID or t=loge X. putting xzet dx2 कर De at or os, [206D-1) + 5D +13450 [ 2D²-2D +5D+1) y zo (2D+3D+1)%20 Auxiliary equation is a am² +3m +Lao am² + 2m +m+lzo em (m+1) 74(M+1)=0 2m +1)(m+1) 20 maod, at 2 Hence, geneed solution is - on2 y = gettqezt y qi(x) + tą (*) / y= q*t tax TW (x=et or, 20 Ang (b) Given , y"-24 + 5y = 3. which 스 be written as cam (D2- 29 +5) y = 3

3 for a2+bx+C =0 xa -b I 562_fac 2X) 2a - 2 ITC 2 Hence, general solution Auxiliary equation is - m² 2m +5=0 m- 27V €2)24x) XS maltz ist them complimentary function, CF = (4 6982X + Es inex) é Now P.I= D² - 2D + 5 D2_2D+5 P-I = O-D 15 .30 (est) 3 ܘܟܘܘ = 3 y=coft P.I y = (4 colure &q8in2x) e + x Ans (C) Given or, y ak² y y" +k key (D9-k+k)420 Auxiliary equation is me-ak² m² + kl=0 (m? - K?j? =0 (m2K²) (m2_K²) =0 m=tk, tk om m= -kiuk, k,k ( Repeated roots) Hence, general solution y = (2+ Qx) etx +G+Gx) ekx where, a, (2, C3 & q are arbitrary constant. k is positive parameter or oo is