Question

If $y = A\cos\theta + B\sin\theta$, then prove that $\dfrac{d^{2}y}{d\theta^{2}} = -y$.

Hint:

Trigonometric functions $\sin\theta$ and $\cos\theta$ always satisfy the differential equation $y'' + y = 0$.

Answer 1

Step 1: Differentiate once. \[ y = A\cos\theta + B\sin\theta \] \[ \dfrac{dy}{d\theta} = -A\sin\theta + B\cos\theta \]

Step 2: Differentiate again. \[ \dfrac{d^{2}y}{d\theta^{2}} = -A\cos\theta - B\sin\theta \]

Step 3: Compare with original function. \[ \dfrac{d^{2}y}{d\theta^{2}} = -(A\cos\theta + B\sin\theta) = -y \]

Final Answer: \[ \boxed{\dfrac{d^{2}y}{d\theta^{2}} = -y} \]