Question

If y = sin (msin^–1x), then which one of the following equations is true?

Options

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

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

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

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

Answer 1

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

Explanation:

Given, y = sin (m(sin^–1x))  ...(i)

Differentiating both sides w.r.t. x, we get

`dy/dx = cos(msin^-1x) xx m/sqrt(1 - x^2)`

`\implies dy/dx = (mcos(msin^-1x))/sqrt(1 - x^2)`  ...(ii)

`\implies y^' = (mcos(msin^-1x))/sqrt(1 - x^2)`  ...(ii)

`\implies (sqrt(1 - x^2))y^' = mcos(msin^-1x)`

Differentiating again w.r.t. 'x', we get

`y^('')(sqrt(1 - x^2)) + y^'((-2x))/(2sqrt(1 - x^2)) = -m^2sin(msin^-1x) 1/sqrt(1 - x^2)`

`\implies` y"(1 – x²) – xy' = – m²y

`\implies` y"(1 – x²) – xy' + m²y = 0

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