Question

If y = Asinx + Bcosx, Where A and B are constants, then \(\frac{d^2y}{dx^2}\) is equal to :

• A. y
• B. -y
• C. x
• D. -x

Answer 1

The Correct Option is B

To find \(\frac{d^2y}{dx^2}\) for \(y = A\sin x + B\cos x\), we first calculate the first derivative \(\frac{dy}{dx}\). Using the derivatives of sine and cosine, we have:

\[\frac{dy}{dx} = A\cos x - B\sin x\]

Next, we differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):

\[\frac{d^2y}{dx^2} = -A\sin x - B\cos x\]

Notice that this result is the negative of the original function \(y\):

\[\frac{d^2y}{dx^2} = -(A\sin x + B\cos x) = -y\]

Therefore, the correct answer is -y.