If f(x)=asin(logx) prove that x^2f"(x)+f'+f(x)=0

f(x) = aSin(Logx)

Differentiating the equation with respect to x:-

f'(x) = a.[ Cos(Logx) ] * (1/x)

x.f'(x) = aCos(Logx)

Differentiating again with respect to x:-

1.f'(x) + x.f''(x) = { a[-Sin(Logx)] } * (1/x)

.

x.f'(x) + (x^2).f''(x) = -aSin(Logx)

.

x.f'(x) + (x^2).f''(x) + aSin(Logx) = 0

.

Here, aSin(Logx)= f(x).

Therefore,

x.f'(x) + (x^2).f''(x) + f(x) = 0

.

Hence proved.

.