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: Direct substitution check:
Substitute \(y=a\cos x+b\sin x\) and its second derivative directly into the left-hand side \(\dfrac{d^2y}{dx^2}+y\).

Step 2: Computing each term separately:
\(\dfrac{d^2y}{dx^2}=-a\cos x-b\sin x\) and \(y=a\cos x+b\sin x\).

Step 3: Adding:
\((-a\cos x-b\sin x)+(a\cos x+b\sin x)=0\) for every \(x\), and for arbitrary constants \(a,b\).

Final Answer:
The equation is satisfied identically.\[ \boxed{\text{Verified: } y''+y=0} \]