Question

7. xy" + 2xy' – 6y = 12 - 16x In x

Answer 1

Let z = ln x => dz/dx = 1/x => dx/dz = x , x = ez

=> dy/dz = (dy/dx)(dx/dz) by chaine rule

= xdy/dx = xy'

d2y/dz2 = d/dx(xy')(dx/dz)

= (y' + xy'') x

= xy' + x2y'' => x2y'' = d2y/dz2 - dy/dz

Now we replace the above terms in the equation

d2y/dz2 - dy/dz + 2dy/dz - 6y = 12 - 16zez

=> Y'' + Y' -6Y = 12 - 16zez , here Y'' = d2y/dz2 , Y' = dy/dz

Corresponding homogeneous equation :

Y'' + Y' - 6Y = 0

if Y = emz be considered as a solution term then auxiliary equation becomes

m2 + m - 6 = 0

=> (m + 3)(m - 2) = 0

=> m = 2,-3

So, complementary function is

Y = y = c1e2z + c2 e-3z

= c1x2 + c2 x-3

Now we find the Particular integral

Y = 1/(D+3)(D-2) {12 - 16zez}

= 12/(0+3)(0-2) e0z - 16ez {1/(D+1+3)(D+1-2)}z

= -2 - 16ez {1/(D+4)(D-1)}z

= -2 -16ez {1/(D2 + 3D -4)}z

= -2 -16ez/(-4) {1 - (D2 + 3D)/4}-1 z

= -2 +4ez [1 + (D2 + 3D)/4 + {(D2 + 3D)/4}2 + . . . . .] z

= -2 + 4ez [z + 0 + 3/4 + 0]

= -2 + 3ez + 4zez

= -2 + 3x + 4 xln x

So, the general solution is

y = c1​​​​​​​x2 + c2 x-3 + [-2 + 3x + 4 xln x]

= -2 + c2 x-3 + 3x + c1​​​​​​​x2 + 4 xlnx [Ans]