Question

Solve the differential equation

e^x tan y dx + (1 + e^x) sec² y dy = 0

Answer 1

e^x tan y dx + (1 + e^x) sec² y dy = 0

∴ e^x tan y dx = – (1 + e^x)sec²ydy

∴ `e^x/(1 + e^x)dx = -(sec^2y)/tany dy`

∴ `int e^x/(1 + e^x)dx = -int (sec^2y)/tany dy`

∴ log (1 + e^x) = – log tan y + log c

∴ log [(1 + e^x)tan y] = log c

∴ (1 + e^x) tan y = c