The differential equation formed by eliminati | vedclass

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(B) Given the equation: $y = A \cos 3x + B \sin 3x$ Differentiating with respect to $x$: $\frac{dy}{dx} = -3A \sin 3x + 3B \cos 3x$ Differentiating ag

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$\frac{d^2 y}{dx^2} + 9y = 0$

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(B) Given the equation: $y = A \cos 3x + B \sin 3x$ Differentiating with respect to $x$: $\frac{dy}{dx} = -3A \sin 3x + 3B \cos 3x$ Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = -9A \cos 3x - 9B \sin 3x$ Factor out $-9$: $\frac{d^2y}{dx^2} = -9(A \cos 3x + B \sin 3x)$ Since $y = A \cos 3x + B \sin 3x$,we substitute $y$ into the equation: $\frac{d^2y}{dx^2} = -9y$ Rearranging gives: $\frac{d^2y}{dx^2} + 9y = 0$

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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