Question

Verify that the function \(y=a\cos x+b\sin x\), in which \(a,b\in R\), is a solution of the differential equation \(\dfrac{d^2y}{dx^2}+y=0\).

Hint:

Differentiate twice and show y'' = −y.

Answer 1

Step 1: First derivative:
\(y=a\cos x+b\sin x \Rightarrow \dfrac{dy}{dx}=-a\sin x+b\cos x\).

Step 2: Second derivative:
\(\dfrac{d^2y}{dx^2}=-a\cos x-b\sin x\).

Step 3: Comparing with y:
Notice \(\dfrac{d^2y}{dx^2}=-(a\cos x+b\sin x)=-y\).

Final Answer:
So \(\dfrac{d^2y}{dx^2}+y=-y+y=0\), confirming \(y=a\cos x+b\sin x\) solves the equation.\[ \boxed{\dfrac{d^2y}{dx^2}+y=0 \text{ holds}} \]