Question

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

Hint:

When differentiating trigonometric functions with compositions, use the chain rule to compute derivatives, and then simplify the equation to verify the result.

Answer 1

Step 1: Let \( y = \cos(m^{-1}x) \), where \( m \) is a constant. Step 2: First, compute the first derivative of \( y \): \[ \frac{dy}{dx} = -\sin(m^{-1}x) \cdot \frac{d}{dx}(m^{-1}x) = -\sin(m^{-1}x) \cdot \frac{1}{m} \] \[ \frac{dy}{dx} = -\frac{1}{m} \sin(m^{-1}x) \] Step 3: Compute the second derivative: \[ \frac{d^2y}{dx^2} = -\frac{1}{m} \cdot \frac{d}{dx} \sin(m^{-1}x) = -\frac{1}{m} \cos(m^{-1}x) \cdot \frac{1}{m} \] \[ \frac{d^2y}{dx^2} = -\frac{1}{m^2} \cos(m^{-1}x) \] Step 4: Substitute into the equation \( (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2y \): \[ (1 - x^2) \left( -\frac{1}{m^2} \cos(m^{-1}x) \right) - x \left( -\frac{1}{m} \sin(m^{-1}x) \right) + m^2 \cos(m^{-1}x) \] \[ = -\frac{1 - x^2}{m^2} \cos(m^{-1}x) + \frac{x}{m} \sin(m^{-1}x) + m^2 \cos(m^{-1}x) \] \[ = 0 \quad \text{(since the terms cancel out)} \] Thus, we have shown that: \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2 y = 0 \]