If y = em sin⁻¹x, then show that (1−x²)d²y/dx² − x dy/dx − m²y = 0

This problem requires the use of chain rule in evaluating derivatives:
y = em sin⁻¹x
dy/dx = em sin⁻¹x . m√(1−x²) = my√(1−x²)
d²y/dx² = d/dx(dy/dx) = m/(1−x²) (√(1−x²).dy/dx + x y/√(1−x²)) = m/(1−x²) (√(1−x²).my√(1−x²) + xy/√(1−x²)) = m/(1−x²) (my + xy/√(1−x²))
⇒ x dy/dx = mxy√(1−x²),
(1−x²)d²y/dx² = m (my + xy/√(1−x²)) = m²y + mxy/√(1−x²)
⇒ (1−x²)d²y/dx² = m²y + x dy/dx
⇒ (1−x²)d²y/dx² − x dy/dx − m²y = 0