The general solution of the differential equation tan ( y ) d x + sec 2 ( y ) · tan ( x ) d y = 0 is

Question

The general solution of the differential equation tan ( y ) d x + sec 2 ( y ) · tan ( x ) d y = 0 is, sin ( y ) · tan ( x ) = c, sin ( x ) · tan ( y ) = c, sin ( x ) + tan ( y ) = c, sin ( x ) - sin ( y ) = c

MARKS App · Accepted Answer

tan y d x + ( sec 2 y ) ( tan x ) d y = 0 d y d x = - tan y sec 2 y tan x On integrating, we get ∫ sec 2 y d y tan y = ∫ - cot x d x ln ( tan y ) = - ln ( sin x ) + c sin x tan y = c 1

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