Question

If \(y = P \cos ux + Q \sin ux\), show that \(\frac{d^2y}{dx^2} + u^2y = 0\).

Hint:

This is a standard differential equation form for simple harmonic motion.
Differentiating trig functions with constants always brings the constant outside the function as a multiplier.

Answer 1

Concept:
• Second-order differentiation.
• Chain rule of differentiation: \(\frac{d}{dx}[\cos(ax)] = -a \sin(ax)\).
• Direct substitution into a differential equation.

Step 1: Find the first derivative of \(y\)
Given: \(y = P \cos ux + Q \sin ux\). Differentiating with respect to \(x\): \[ \frac{dy}{dx} = P(-u \sin ux) + Q(u \cos ux) \] \[ \frac{dy}{dx} = -Pu \sin ux + Qu \cos ux \]

Step 2: Find the second derivative of \(y\)
Differentiating again with respect to \(x\): \[ \frac{d^2y}{dx^2} = -Pu(u \cos ux) + Qu(-u \sin ux) \] \[ \frac{d^2y}{dx^2} = -Pu^2 \cos ux - Qu^2 \sin ux \]

Step 3: Substitute into the differential expression and prove
Factor out \(-u^2\) from the expression: \[ \frac{d^2y}{dx^2} = -u^2(P \cos ux + Q \sin ux) \] Since \(P \cos ux + Q \sin ux = y\): \[ \frac{d^2y}{dx^2} = -u^2y \] Rearranging terms: \[ \frac{d^2y}{dx^2} + u^2y = 0 \] Hence proved.