Question

For the equation xạy" +6xy' +9y=0, the general solution is y=Cqx+3 +C2x-3in(x) None of the other alternatives is correct. y=Cje-3+Czte-37 y= C1x-5/2 cos(711x/2) + C2x-5/2sin(711x/2). v=C1x1-5+V11) V11)/2 + C2xl-5-v11)/2

Answer 1

I am attaching the pictures of solution. Thanks

Ang x ²4" + 6x Y' + gy=0 bet x= ez -) enx = ln (62) U lax=2 (as en (69) = a =) 2= lnx - ③ 5) dz da =d (enx) da d (lnx) = 1 2) as dz doo 1 х 14 Now dy 11 da dy dz dz dy d2 X - ) dy da =

ولے ( I dy y = dy da da x da, dx - x dz 2 x dy dy doc dz There = xy = Dy D = d dz Now day d dy I dy 2 x y = DY - o dx2 dxda dey (① ha راے dy x dal dz ( t dy d dz delo)

C 1 dl dy x dal az dz + dy d (x²) da dz do I dy х d 2² (+) dz + dy (-1,82 (using ] as & (x") = mx"- da x² dze dx² d'y = I dy dy dx2 xe da = x2 dzy dy dy dz² dz There y'= dy =) x²y" - D²y- DY ď (y) = D²y = D(D-1) y but 6, 2 in ① D (D-1) y +6 Dy + 99= dx² =) x²y" d2² -o

-0 - 0 (0(0-1) +6D+ 1 700D +60 +3] ya - (0+50 +9) y auxiliary equation will be:- ma +5m +9=0 > m= -5 ± √5)2_4 (1) (9) 211) if ama because tom+c=0 then m=-bt ✓ Gayac la Here a=1,6= 5, C=9 -) m = -5 + √25-36 lo

m= .57-1 2 = m = -5+ i si where it g = m = -3/5 I I so general solution will be:- We know that it m- &t iß then Yo ed" [c, Cos (0-2) +£,3in (82) here 2=-2 / 4 B=51

so و) sin y = 2 & 3 [c, Cos/ 5 1 1 2 + Y = (22) 3 (C, cos (517) + Csin Zte sin luz y - 5/2 X {a C. Cos (VII enx) + (asin (1 box) (using @ and ) As this option is not given. So None of the other alternativers is correct.) (second option is correct.) answer will be