If y=A cos n x+B sin n x, then y_2+n^2 y is e | vedclass

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(A) Given the function: $y = A \cos n x + B \sin n x$. First, differentiate with respect to $x$: $y_1 = \frac{dy}{dx} = -A n \sin n x + B n \cos n x$.

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(A) Given the function: $y = A \cos n x + B \sin n x$. First, differentiate with respect to $x$: $y_1 = \frac{dy}{dx} = -A n \sin n x + B n \cos n x$. Now, differentiate again with respect to $x$ to find the second derivative $y_2$: $y_2 = \frac{d^2y}{dx^2} = -A n^2 \cos n x - B n^2 \sin n x$. Factor out $-n^2$: $y_2 = -n^2 (A \cos n x + B \sin n x)$. Since $y = A \cos n x + B \sin n x$, we can substitute $y$ into the equation: $y_2 = -n^2 y$. Rearranging the terms gives: $y_2 + n^2 y = 0$.

If $y=A \cos n x+B \sin n x$, then $y_2+n^2 y$ is equal to

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