Question

If \(y=5\cos x-3\sin x\), then prove that \(\dfrac{d^2y}{dx^2}+y=0\).

Hint:

Differentiate twice and notice the second derivative is exactly -y.

Answer 1

Step 1: Differentiate once:
\[ \dfrac{dy}{dx}=-5\sin x-3\cos x \]

Step 2: Differentiate again:
\[ \dfrac{d^2y}{dx^2}=-5\cos x+3\sin x=-(5\cos x-3\sin x)=-y \]

Step 3: Rearrange:
\[ \dfrac{d^2y}{dx^2}+y=-y+y=0 \]

Final Answer:
\[ \boxed{\dfrac{d^2y}{dx^2}+y=0} \]