A) \[xy\,\sec x=\tan x+c\]
B) \[xy\,\tan x=\sec \,x+c\]
C) \[y\,\sec x=x\tan x+c\]
D) \[y\tan x=x\sec x+c\]
Correct Answer: A
Solution :
[a] We have \[\frac{dy}{dx}+y\left( \tan x+\frac{1}{x} \right)=\frac{1}{x\cos x}\] \[I.F.={{e}^{\int{\left( \tan x\frac{1}{x} \right)\,dx}}}=x\,\sec x\] Therefore, solution is: \[y\times (x\sec x)=\int{\frac{1}{x\cos x}}\times (x\sec x)\,dx\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,xy\sec x=\tan x+c\]
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