A) \[y=\tan x-1+c{{e}^{-\tan x}}\]
B) \[{{y}^{2}}=\tan x-1+c{{e}^{\tan x}}\]
C) \[y{{e}^{\tan x}}=\tan x-1+c\]
D) \[y{{e}^{-\tan x}}=\tan x-1+c\]
Correct Answer: A
Solution :
I.F. = \[{{e}^{\int{{{\sec }^{2}}x\,dx}}}={{e}^{\tan x}}\] \ Solution is \[y{{e}^{\tan x}}=c+\int{\tan x{{e}^{\tan x}}{{\sec }^{2}}x\,dx}\] Þ \[y=c{{e}^{-\tan x}}+\tan x-1\].You need to login to perform this action.
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