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If $y=5 \cos x-3 \sin x$, prove that $\frac{d^2 y}{d x^2}+y=0$
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Verified Answer
$\frac{d y}{d x}=-5 \sin x-3 \cos x$
$\frac{d^2 y}{d x^2}=-5 \cos x+3 \sin x=-y$
$\Rightarrow \frac{d^2 y}{d x^2}+y=0$. Hence proved.
$\frac{d^2 y}{d x^2}=-5 \cos x+3 \sin x=-y$
$\Rightarrow \frac{d^2 y}{d x^2}+y=0$. Hence proved.
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