Solution of the differential equation \[\frac{dy}{dx}+y{{\sec }^{2}}x=\tan x{{\sec }^{2}}x\] is [DCE 2001, 05]

\[{{y}^{2}}=\tan x-1+c{{e}^{\tan x}}\]

\[y=\tan x-1+c{{e}^{-\tan x}}\]

\[y{{e}^{-\tan x}}=\tan x-1+c\]

Solution of the differential equation \[\frac{dy}{dx}+y{{\sec }^{2}}x=

1.	Solution of the differential equation \[\frac{dy}{dx}+y{{\sec }^{2}}x=\tan x{{\sec }^{2}}x\] is [DCE 2001, 05]
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.	\[y{{e}^{-\tan x}}=\tan x-1+c\]
Answer» B. \[{{y}^{2}}=\tan x-1+c{{e}^{\tan x}}\]

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Solution of the differential equation \[\frac{dy}{dx}+y{{\sec }^{2}}x=\tan x{{\sec }^{2}}x\] is [DCE 2001, 05]

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