Question

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

Hint:

When showing differential equations from trig functions, the second derivative usually results in the original function multiplied by a constant square of the frequency (\( -u^2 \)).

Answer 1

Concept:
• Successive differentiation involves differentiating a function multiple times.
• Chain rule: \( \frac{d}{dx}(\cos ax) = -a \sin ax \) and \( \frac{d}{dx}(\sin ax) = a \cos ax \).

Step 1: Find the first derivative \( \frac{dy}{dx} \)
Differentiate \( y \) 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 \( \frac{d^2y}{dx^2} \)
Differentiate \( \frac{dy}{dx} \) 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: Simplify and relate to \( y \)
Factor out \( -u^2 \) from the terms on the right-hand side:
\[ \frac{d^2y}{dx^2} = -u^2(P \cos ux + Q \sin ux) \]
Since the term in parentheses is original function \( y \):
\[ \frac{d^2y}{dx^2} = -u^2y \implies \frac{d^2y}{dx^2} + u^2y = 0 \]
Hence proved.