Question

If \( y = 2 \sin x + 3 \cos x \) and \( y + A \frac{d^2 y}{dx^2} = B \), then the values of \( A, B \) are respectively

• A. \(0, 1\)
• B. \(0, -1\)
• C. \(-1, 0\)
• D. \(1, 0\)

Hint:

For a linear combination of \(\sin x\) and \(\cos x\), the second derivative is the negative of the original function. This property makes it easy to determine \(A\) and \(B\) by comparing coefficients.

Answer 1

The Correct Option is D

Step 1: Understanding the Question: We are given \(y = 2\sin x + 3\cos x\) and the relation \(y + A \frac{d^2 y}{dx^2} = B\). We need to find constants \(A\) and \(B\).

Step 2: Key Formula or Approach: Compute the first and second derivatives of \(y\), then substitute into the given equation and equate coefficients for all \(x\).

Step 3: Detailed Explanation: First derivative: \(\frac{dy}{dx} = 2\cos x - 3\sin x\). Second derivative: \(\frac{d^2 y}{dx^2} = -2\sin x - 3\cos x = -(2\sin x + 3\cos x) = -y\). Now substitute into \(y + A \frac{d^2 y}{dx^2} = B\): \[ y + A(-y) = B \quad\Rightarrow\quad y(1 - A) = B. \] This equality must hold for all \(x\). The left side is a function of \(x\) unless the coefficient of \(y\) is zero; otherwise it would vary with \(x\) while the right side is constant. Therefore, we require \(1 - A = 0\) and then \(B = 0\). Thus \(A = 1\) and \(B = 0\).

Step 4: Final Answer: \(A = 1, B = 0\), which corresponds to option (D).