A) \[y=\sin x+{{c}_{1}}x+{{c}_{2}}\]
B) \[y=\cos x+{{c}_{1}}x+{{c}_{2}}\]
C) \[y=\tan x+{{c}_{1}}x+{{c}_{2}}\]
D) \[y=\log \sin x+{{c}_{1}}x+{{c}_{2}}\]
Correct Answer: A
Solution :
We have , \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+\sin x=0\] or \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\sin x\] On integrating, \[\frac{dy}{dx}=-\left( -\cos x \right)+{{c}_{1}}=\cos x+{{c}_{1}}\] Again integrate, we get\[y=\sin x+{{c}_{1}}x+{{c}_{2}}\].
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