A) \[y=\log \cos x+cx\]
B) \[y=\log \sec x+{{c}_{1}}x+{{c}_{2}}\]
C) \[y=\log \sec x-{{c}_{1}}x+{{c}_{2}}\]
D) None of these
Correct Answer: B
Solution :
\[{{\cos }^{2}}x\frac{{{d}^{2}}y}{d{{x}^{2}}}=1\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{\sec }^{2}}x\] On integrating, we get \[\frac{dy}{dx}=\tan x\pm {{c}_{1}}\] Again integrating, we get \[y=\log \sec x\pm {{c}_{1}}x\pm {{c}_{2}}\].
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