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

Explanation

`y = acos(logx)+bsin(logx)`

Let `logx = t =>1/xdx = dt=>dt/dx = 1/x`

Then, `y = acost+bsint`

`=>dy/dt = -asint+bcost`

`:.dy/dx = dy/dt*dt/dx = (-asint+bcost)1/x`

`=>xdy/dx = bcost-asint`

Differentiating w.r.t. `x`,

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

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