Question

If x = cos t, y = cos mt, prove that `(1 - x^2) (d^2y)/(dx^2) - x (dy)/(dx) + m^2y = 0`.

Answer 1

⇒ Given: x = cos t, y = cos mt,

And we know that,

`(dx)/(dt) = - sin t`

`(dy)/(dt) = - m sin mt`

⇒ So, `(dy)/(dx) = ((dy)/(dt))/((dx)/(dt))`

`(dy)/(dx) = (- m sin mt)/(- sin t)`

`(dy)/(dx) = (m sin mt)/(sin t)`

⇒ Now, (1 − x²) = 1 − cos² t = sin² t

∴ `(1 - x^2) (dy)/(dx) = sin^2 t * (m sin mt)/(sin t)`

`(1 - x^2) (dy)/(dx) = m sin t sin mt`

⇒ Differentiate with respect to x:

`d/(dx)[(1 - x^2)(dy)/(dx)] = d/(dx) (m sin t sin mt)`

⇒ Using: `d/(dx) = 1/((dx)/(dt)) * d/(dt) = -1/(sin t) * d/(dt)`

`d/(dx) (m sin t sin mt) = -1/(sin t) * d(dt)(m sin t sin mt)`

= `-m/(sin t) (cos t sin mt + m sin t cos mt)`

= `-m (cos t sin mt)/(sin t) - m^2 cos mt`

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

∴ `d/(dx)[(1 - x^2)(dy)/(dx)] = - x (dy)/(dx) - m^2y`

⇒ But, `d/(dx)[(1 - x^2)(dy)/(dx)] = (1 - x^2)(d^2y)/(dx^2) - 2x (dy)/(dx)`

⇒ So,

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

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

Hence proved.