Question

If \( y = \cos (m \cos^{-1} x) \), then show that \( (1 - x^2) \frac{d^2 y{dx^2} - x \frac{dy}{dx} + m^2 y = 0 \).
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Hint:

For trigonometric differential equations, use chain rule and simplify using identities.

Answer 1

Let \( \theta = \cos^{-1} x \), so \( x = \cos \theta \), \( y = \cos m\theta \).
\[ \frac{dx}{d\theta} = -\sin \theta, \quad \frac{d\theta}{dx} = -\frac{1}{\sin \theta} = -\frac{1}{\sqrt{1 - x^2}}. \] \[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} = (-m \sin m\theta) \cdot \left( -\frac{1}{\sqrt{1 - x^2}} \right) = \frac{m \sin m\theta}{\sqrt{1 - x^2}}. \] Second derivative: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{m \sin m\theta}{\sqrt{1 - x^2}} \right). \] Use product rule: Let \( u = m \sin m\theta \), \( v = (1 - x^2)^{-1/2} \).
\[ \frac{du}{dx} = m \cdot m \cos m\theta \cdot \frac{d\theta}{dx} = \frac{-m^2 \cos m\theta}{\sqrt{1 - x^2}}, \quad \frac{dv}{dx} = -\frac{1}{2} (1 - x^2)^{-3/2} \cdot (-2x) = \frac{x}{(1 - x^2)^{3/2}}. \] \[ \frac{d^2 y}{dx^2} = \frac{\frac{-m^2 \cos m\theta}{\sqrt{1 - x^2}} \cdot \sqrt{1 - x^2} - m \sin m\theta \cdot \frac{x}{(1 - x^2)^{3/2}}}{1 - x^2}. \] \[ = \frac{-m^2 \cos m\theta - \frac{m x \sin m\theta}{1 - x^2}}{1 - x^2} = \frac{-m^2 \cos m\theta (1 - x^2) - m x \sin m\theta}{(1 - x^2)^2}. \] Left-hand side: \[ (1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + m^2 y = (1 - x^2) \cdot \frac{-m^2 \cos m\theta (1 - x^2) - m x \sin m\theta}{(1 - x^2)^2} - x \cdot \frac{m \sin m\theta}{\sqrt{1 - x^2}} + m^2 \cos m\theta. \] \[ = \frac{-m^2 \cos m\theta (1 - x^2) - m x \sin m\theta}{1 - x^2} - \frac{m x \sin m\theta}{\sqrt{1 - x^2}} \cdot \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}} + m^2 \cos m\theta. \] \[ = -m^2 \cos m\theta - \frac{m x \sin m\theta}{1 - x^2} - \frac{m x \sin m\theta}{1 - x^2} + m^2 \cos m\theta = 0. \] Proved.