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

The general solution of the differential equation tan(y)dx+sec2(y)·tan(x)dy=0 is
  1. sin(y)·tan(x)=c
  2. sin(x)·tan(y)=c
  3. sin(x)+tan(y)=c
  4. sin(x)-sin(y)=c

Solution

tanydx+(sec2y)(tanx)dy=0

dydx=-tanysec2ytanx

On integrating, we get

∫sec2ydytany=∫-cotxdx

ln(tany)=-ln(sinx)+c

sinxtany=c1

Asked in: AP EAMCET 2020 (23 Sep Shift 1)

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