Question

If `y = (x + sqrt(a^2 + x^2))^m`, prove that `(a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`

Answer 1

`y = [x + sqrt(a^2 + x^2)]^m`

`dy/dx = (m[x + sqrt(a^2 + x^2)]^m)/(x + sqrt(a^2 + x^2))*[1 + (1*2x)/(2sqrt(a^2 + x^2))]`

= `my 1/sqrt(a^2 + x^2)`

`sqrt(a^2 + x^2)dy/dx = my`

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

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

`(a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`

Answer 2

`y = [x + sqrt(a^2 + x^2)]^m`

`dy/dx = (m[x + sqrt(a^2 + x^2)]^m)/(x + sqrt(a^2 + x^2)]*[1 + (1.2x)/(2sqrt(a^2 + x^2))]`

= `my 1/sqrt(a^2 + x^2)`

`sqrt(a^2 + x^2)dy/dx = my`

Squaring both sides we get,

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

Differentiating w.r.t ‘𝑥’,

`2x(dy/dx)^2 + (a^2 + x^2)2dy/dx*(d^2y)/(dx^2) = 2m^2ydy/dx`

`\implies (a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`