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: Read the question.
Given $y = 2\sin x + 3\cos x$ and the relation $y + A\,\dfrac{d^2y}{dx^2} = B$, find $A$ and $B$.
Step 2: First derivative.
\[ \frac{dy}{dx} = 2\cos x - 3\sin x \]
Step 3: Second derivative.
\[ \frac{d^2y}{dx^2} = -2\sin x - 3\cos x = -(2\sin x + 3\cos x) = -y \] So the second derivative is just $-y$.
Step 4: Substitute into the relation.
\[ y + A(-y) = B \;\Rightarrow\; y(1 - A) = B \]
Step 5: Make it hold for all $x$.
The left side changes with $x$ unless its coefficient is zero, but the right side $B$ is a fixed number. So we need $1 - A = 0$, which forces $A = 1$, and then $B = 0$.
Step 6: State the values.
\[ A = 1,\quad B = 0 \] \[ \boxed{A = 1,\ B = 0} \]