Question

Solve
$3e^{x} tan y dx + (1+e^{x}) sec^{2} y dy$

Answer 1

3 e^x tan y  dx + (1+e^x) sec^2 y  dy = 0

$-3\frac{e^x}{1+e^x} dx =\frac{sec^2y}{tany}dy=2\ cosec\ 2y\ dy\\\\integrate\ both\ sides:\\\\-3Ln(1+e^x)-Ln K=-2\ Ln | cosec\ 2y+cot\ 2y |\\\\K(1+e^x)^3=(cosec\ 2y+cot2y)^2=(\frac{1+cos2y}{sin2y} )^2\\\\cot^2y=K(1+e^x)^3$

K, K1 are the integration constants.

y = Tan⁻¹ [ K1 (1+eˣ)⁻³/² ]