Question

If `x = tan (1/a log y)` then show that `(1 + x^2) (d^2y)/(dx^2) + (2x - a) dy/dx = 0`.

Answer 1

Given, `x = tan (1/a log y)`

a tan^–1x = log y

⇒ `y = e^(a tan^-1x)`

Differentiate w.r.t. x on both sides

⇒ `dy/dx = e^(a tan^-1x) a/(1 + x^2)`   ...(i)

⇒ `(1 + x^2) dy/dx = ae^(a tan^-1x)`

Again, differentiate w.r.t. x on both sides

`(1 + x^2) (d^2y)/(dx^2) + 2x dy/dx = ae^(a tan^-1x) * a/(1 + x^2)`

From equation (i),

`(1 + x^2) (d^2y)/(dx^2) + 2x dy/dx = a dy/dx`

⇒ `(1 + x^2) (d^2y)/(dx^2) + (2x - a) dy/dx = 0`

Hence Proved.