Question

If y = `(x + sqrt(x^2 - 1))^m`, show that `(x^2 - 1)(d^2y)/(dx^2) + xdy/dx` = m²y

Answer 1

Given, y = `(x + sqrt(x^2 - 1))^m`   ...(1)

Differentiating w.r.t. x,

`dy/dx = m(x + sqrt(x^2 - 1))^(m - 1) d/dx(x + sqrt(x^2 - 1))`

∴ `dy/dx = m(x + sqrt(x^2 - 1))^(m - 1){1 + x/sqrt(x^2 - 1)}` 

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

∴ `dy/dx = (my)/sqrt(x^2 - 1)`   ...[From (1)]

∴ `sqrt(x^2 - 1) dy/dx` = my

Squaring both sides,

`(x^2 - 1)(dy/dx)^2` = m²y²

Again differentiating w.r.t. x,

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

Dividing both sides by `(2dy)/dx`

`(x^2 - 1)(d^2y)/(dx^2) + xdy/dx` = m²y