Question

If `y = e^(msin^-1x)`, prove that `(1 - x^2) (d^2y)/(dx^2) - x dy/dx = m^2y`

Answer 1

Given: `y = e^(msin^-1x)`

Differentiating w.r.t. x, we get

`dy/dx = e^(msin^-1x) d/dx (m sin^-1x)`

⇒ `dy/dx = y xx m/sqrt(1 - x^2)`   ...(i)

or `sqrt(1 - x^2) dy/dx = my`

Squaring both LHS and RHS,

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

Differentiating w.r.t. x,

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

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

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

Hence Proved.