3ex tan y dx + (1 + ex) sec2 y dy = 0, when x = 0, y = .
$(3"e"^"x")/(1 + "e"^x) "dx" + ("sec"^2"y")/("tan y") "dy" = 0$
Integrating both sides, we get
$3 int "e"^"x"/(1 + "e"^"x") "dx" + int (sec^2"y")/(tan "y") "dy" = "c"_1$
Each of these integrals is of the type
$int ("f"'("x"))/("f"("x")) "dx" = log |"f"(x)| + "c"$
the general solution is
3 log |1 + ex| + log |tan y| = log c, where c1 =log c
log |(1 + ex)3 * tan y| = log c
(1 + ex)3 tan y = c
When x = 0, y = , we have
(1 + e0)3 tan = c
c = 0
the particular solution is (1 + ex)3 tan y = 0
