A) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]
B) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-y=0\]
C) \[\frac{dy}{dx}+y=0\]
D) None of these
Correct Answer: A
Solution :
\[y=A\sin x+B\cos x\]Þ \[\frac{dy}{dx}=A\cos x-B\sin x\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-A\sin x-B\cos x\]\[=-(A\sin x+B\cos x)=-y\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]is the required differential equation.
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