Question

If `y= sinx/ x^2`, find `dy/dx` and `(d^2y)/(dx^2)`. Then prove that `x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y = 0` ?

Answer 1

We seek to show that:

`x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y = 0` where `y= sinx/x^2`

Using the quotient rule then differentiating `y= sinx/x^2` wrt `x` we have:

`dy/dx = { (x^2)(cosx)-(2x)(sinx) } / (x^2)^2`
`\ \ \ \ \ \ = (xcosx-2sinx)/x^3`

And differentiating a second time, we have:

`(d^2y)/(dx^2) = { (x^3)(-xsinx+cosx-2cosx) - (3x^2)(xcosx-2sinx) } / (x^3)^2`

`\ \ \ \ \ \ \ = { -x^2sinx+xcosx-2xcosx - 3xcosx+6sinx) } / (x^4)`

`\ \ \ \ \ \ \ = ( -x^2sinx-4xcosx +6sinx ) / (x^4)`

And so, considering the LHS of the given expression:

`LHS = x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y`

`\ \ \ \ \ \ \ \ = x^2 {( -x^2sinx-4xcosx +6sinx ) / (x^4)}`
`\ \ \ \ \ \ \ \ \ \ \ \ + 4x {(xcosx-2sinx)/x^3}`
`\ \ \ \ \ \ \ \ \ \ \ \ + (x^2 + 2) {sinx/x^2}`

`\ \ \ \ \ \ \ \ = {( -x^2sinx-4xcosx +6sinx ) / (x^2)}`
`\ \ \ \ \ \ \ \ \ \ \ \ + 4 {(xcosx-2sinx)/x^2}`
`\ \ \ \ \ \ \ \ \ \ \ \ + (x^2 + 2) {sinx/x^2}`

`\ \ \ \ \ \ \ \ = 1/x^2 {-x^2sinx-4xcosx +6sinx + 4xcosx-8sinx + (x^2 + 2)sinx}`

`\ \ \ \ \ \ \ \ = 0 \ \ \` QED