Question

The differential equation formed by eliminating arbitrary constants \( A \) and \( B \) from the equation \[ y = A \cos 3x + B \sin 3x \] is:

• A. \( \frac{d^2y}{dx^2} + y = 0 \)
• B. \( \frac{d^2y}{dx^2} + 9y = 0 \)
• C. \( \frac{d^2y}{dx^2} - 9y = 0 \)
• D. \( \frac{d^2y}{dx^2} - y = 0 \)

Hint:

When dealing with trigonometric functions, use standard differentiation rules and then simplify the equation to eliminate arbitrary constants.

Answer 1

The Correct Option is B

Step 1: Differentiate the given equation The given equation is: \[ y = A \cos 3x + B \sin 3x \] First, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = -3A \sin 3x + 3B \cos 3x \] Now differentiate again to get the second derivative: \[ \frac{d^2y}{dx^2} = -9A \cos 3x - 9B \sin 3x \] Step 2: Eliminate \( A \) and \( B \) Notice that: \[ \frac{d^2y}{dx^2} = -9y \] Thus, the differential equation is: \[ \frac{d^2y}{dx^2} + 9y = 0 \]

Answer 2

Given the general solution:
\[ y = A \cos 3x + B \sin 3x \] where \( A \) and \( B \) are arbitrary constants.

Step 1: Differentiate \( y \) twice with respect to \( x \):
\[ \frac{dy}{dx} = -3A \sin 3x + 3B \cos 3x \] \[ \frac{d^2y}{dx^2} = -9A \cos 3x - 9B \sin 3x \]

Step 2: Substitute back into the expression for \( y \):
\[ \frac{d^2y}{dx^2} = -9 (A \cos 3x + B \sin 3x) = -9y \] Rearranged:
\[ \frac{d^2y}{dx^2} + 9y = 0 \]

Therefore, the differential equation formed by eliminating \( A \) and \( B \) is:
\[ \boxed{\frac{d^2y}{dx^2} + 9y = 0} \]