Question

If `y = (tan^(-1)x)^2` then show that `(1+x^2)^2y'' + 2x (1+x^2) y' = 2`?

Answer 1

To proceed we will need some standard Calculus results:

`d/dx tan^(-1)x = 1/(1+x^2)`

Now we have:

`y = (tan^(-1)x)^2`

If we apply the chain rule then we get:

`y' = 2*(tan^(-1)x)*1/(1+x^2)`
`\ \ \ \ = (2tan^(-1)x)/(1+x^2)`

And differentiating again and applying the quotient rule, along with the chain rule, we get:

`y'' = { (1+x^2)(d/dx 2tan^(-1)x) - (2tan^(-1)x)(d/dx (1+x^2)) } / (1+x^2)^2`

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

`\ \ \ \ = { 2 - 2x * (2tan^(-1)x) } / (1+x^2)^2`

`:. (1+x^2)^2y'' = 2 - 2x * (2tan^(-1)x)`

`:. (1+x^2)^2y'' = 2 - 2x (1+x^2)* (2tan^(-1)x)/(1+x^2)`

`:. (1+x^2)^2y'' = 2 - 2x (1+x^2) y'`

`:. (1+x^2)^2y'' + 2x (1+x^2) y' = 2` QED