Question

Show that if `y= (sin^-1 x)^2` then `(1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2 = 0` ?

Answer 1

We seek to show that:

`(1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2 = 0` where `y= (sin^-1 x)^2`

Using the result:

`d/dx(arcsin x) = 1/sqrt(1-x^2)`

In conjunction with the chain rule, then differentiating `y= (sin^-1 x)^2` wrt `x` we have:

`dy/dx = (2arcsinx)/sqrt(1-x^2)`

And differentiating a second time, in conjunction with the quotient rule, we have:

`(d^2y)/(dx^2) = { (sqrt(1-x^2))(2/sqrt(1-x^2)) - ((1/2 (-2x))/sqrt(1-x^2))(2arcsinx) } / (sqrt(1-x^2))^2`

`\ \ \ \ \ \ \ = ( 2 + (2x \ arcsinx)/sqrt(1-x^2) ) / (1-x^2)`

And so, considering the LHS of the given expression:

`LHS = (1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2`

`\ \ \ \ \ \ \ \ = (1 - x^2) {( 2 + (2x \ arcsinx)/sqrt(1-x^2) ) / (1-x^2)} - x{(2arcsinx)/sqrt(1-x^2)} - 2`

`\ \ \ \ \ \ \ \ = 2 + (2x \ arcsinx)/sqrt(1-x^2) - (2xarcsinx)/sqrt(1-x^2) - 2`

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