If `y="sin"(logx),` then prove that `(x^2d^2y)/(dx^2)+x(dy)/(dx)+y=0`

Explanation

`y = sin(logx)`

`dy/dx = cos(logx)*1/x`

`=>xdy/dx = cos(logx)`

Again differentiating w.r.t. `x`,

`=>x(d^2y)/dx^2 + dy/dx = -sin(logx)*1/x`

`=>x(d^2y)/dx^2+dy/dx = -y/x`

`=>x^2(d^2y)/dx^2+xdy/dx+y = 0`