Question:

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

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When dealing with trigonometric functions, use standard differentiation rules and then simplify the equation to eliminate arbitrary constants.
Updated On: Mar 24, 2025
  • \( \frac{d^2y}{dx^2} + y = 0 \)
  • \( \frac{d^2y}{dx^2} + 9y = 0 \)
  • \( \frac{d^2y}{dx^2} - 9y = 0 \)
  • \( \frac{d^2y}{dx^2} - y = 0 \)
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The Correct Option is B

Solution and Explanation

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: \[ \frac{d^2y}{dx^2} + 9y = 0 \]
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