It is given that, y = 3cos(log x) + 4sin(log x)
Then,

dy

dx

= 3 ×

d

dx

[cos log x] + 4 ×

d

dx

[sin log x]
= 3 × −sinlogx×

d

dx

logx] + 4 × [coslogx×

d

dx

logx]
=

−3sinlogx

x

+

4coslogx

x

=

4coslogx−3sinlogx

x

d2y

dx2

=

d

dx

(

4cos(logx)−3sin(logx)

x

)
=
=
=
=
∴ x2

d2y

dx2

+ x

dy

dx

+ y = x2(

−sin(logx)−7cos(logx)

x2

) + x (

4cos(logx)−3sin(logx)

x

) + 3 cos (log x) + 4 sin (log x)
= - sin (log x) - 7 cos (log x) + 4 cos (log x) - 3 sin (log x) + 3 cos (log x) + 4 sin (log x)
= 0
Hence proved.