A) \[\sin x+{{c}_{1}}x+{{c}_{2}}\]
B) \[\cos x+{{c}_{1}}x+{{c}_{2}}\]
C) \[\tan x+{{c}_{1}}x+{{c}_{2}}\]
D) \[\log \sin x+{{c}_{1}}x+{{c}_{2}}\]
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, \[\frac{dy}{dx}=-(-\cos x)+{{c}_{1}}\] = \[\cos x+{{c}_{1}}\] Again integrate, we get \[y=\sin x+{{c}_{1}}x+{{c}_{2}}\].
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