Question

If y = (tan^–1 x)², show that (x² + 1)² y₂ + 2x (x² + 1) y₁ = 2

Answer 1

Let y = (tan^–1 x)²   ....(1)

Differentiating (1) w.r.t. x, we get,

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

`(d^2y)/dx^2 = 2 [tan^-1 x (({1 + x^2} . 0 - 2x))/(1 + x^2)^2 + 1/ (1 + x^2). 1/ (1 + x^2)]`

= `2 [(-2x tan^-1 x)/ (1 +x^2)^2 + 1/ (1 + x^2)^2]`

= `2 [(-2x tan^-1 x + 1)/ (1 + x^2)^2]`

Now, `(x^2 + 1)^2 (d^2y)/dx^2 + 2x (x^2 + 1) dy/dx`

= `(x^2 + 1)^2 . 2 [(-2x tan^-1x + 1)/ (1 + x^2)^2] + 2x (x^2 + 1). 2tan ^-1 x. 1/ (1 + x^2)`

= −4x tan⁻¹ x + 2 + 4x tan⁻¹ x

= 2