Question

If $y = 5 \cos x - 3 \sin x, \text{ prove that } \frac{d^2y}{dx^2} + y = 0$.

Hint:

Quick Tip: When proving second-order derivatives, make sure to take the derivative twice and simplify. The goal is often to eliminate terms and show that the equation holds.

Answer 1

Given \( y = 5 \cos x - 3 \sin x \). The first derivative with respect to \( x \) is calculated as: \[ \frac{dy}{dx} = \frac{d}{dx} \left( 5 \cos x - 3 \sin x \right) \] Applying standard derivatives \( \frac{d}{dx} (\cos x) = -\sin x \) and \( \frac{d}{dx} (\sin x) = \cos x \): \[ \frac{dy}{dx} = -5 \sin x - 3 \cos x \] The second derivative, \( \frac{d^2y}{dx^2} \), is found by differentiating \( \frac{dy}{dx} \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( -5 \sin x - 3 \cos x \right) \] Using the derivatives \( \frac{d}{dx} (\sin x) = \cos x \) and \( \frac{d}{dx} (\cos x) = -\sin x \): \[ \frac{d^2y}{dx^2} = -5 \cos x + 3 \sin x \] Finally, the sum of \( \frac{d^2y}{dx^2} \) and \( y \) is computed: \[ \frac{d^2y}{dx^2} + y = \left( -5 \cos x + 3 \sin x \right) + \left( 5 \cos x - 3 \sin x \right) \] Simplifying the expression: \[ = (-5 \cos x + 5 \cos x) + (3 \sin x - 3 \sin x) \] \[ = 0 \] This demonstrates that: \[ \frac{d^2y}{dx^2} + y = 0 \]