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: First derivative term by term:
$\dfrac{d}{dx}(5\cos x)=-5\sin x$, $\dfrac{d}{dx}(-3\sin x)=-3\cos x$. Sum: $y'=-5\sin x-3\cos x$.

Step 2: Second derivative term by term:
$\dfrac{d}{dx}(-5\sin x)=-5\cos x$, $\dfrac{d}{dx}(-3\cos x)=3\sin x$. Sum: $y''=-5\cos x+3\sin x$.

Step 3: Compare y'' to -y:
$-y=-(5\cos x-3\sin x)=-5\cos x+3\sin x$, which is identical to $y''$ found above.

Final Answer:
$y''=-y \Rightarrow y''+y=0$. \[ \boxed{y''+y=0} \]