$("e"^"y" + 1) cos "x" + "e"^"y" sin "x" "dy"/"dx" = 0$
$(cos "x")/(sin "x") "dx" + "e"^"y"/("e"^"y" + 1) "dy" = 0$
Integrating both sides, we get
$int (cos "x")/(sin "x") "dx" + int "e"^"y"/("e"^"y" + 1) "dy" = "c"_1$
Now, $"d"/"dx"(sin "x") = cos "x", "d"/"dx" ("e"^"y" + 1) = "e"^"y"$ and
$int ("f"'("x"))/("f"("x"))$ dx = log |f(x)| + c
from (1), the general solution is
log |sin x| + log |ey + 1| = log c, where c1 = log c
$sin "x" * ("e"^"y" + 1) = "c"$
When x = $pi/4$, y = 0, we get
$(sin pi/4)("e"^0 + 1) = "c"$
c = $1/sqrt2 (1 + 1) = sqrt2$
the particular solution is sin x$* ("e"^"y" + 1) = sqrt2$
