$("x" + 1) "dy"/"dx" - 1 = 2"e"^-"y"$
$("x" + 1) "dy"/"dx" = 2/"e"^"y" + 1 = (2 + "e"^-"y")$
$"e"^"y"/(2 + "e"^-"y") "dy" = 1/("x" + 1)$dx
Integrating both sides, we get
$int "e"^"y"/(2 + "e"^-"y") "dy" = int 1/("x" + 1)$dx
log |2 + ey| = log |x + 1| + log c ......$[ "d"/"dy" (2 + "e"^"y") = "e"^"y" and int("f"'("y"))/("f"("y")) "dy" = log |"f"("y")| + "c"]$
log |2 + ey| = log |c (x + 1)|
2 + ey = c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
2 + e0 = c (1 + 1)
3 = 2c
c = $3/2$
the particular solution is $2 + "e"^"y" = 3/2("x" + 1)$
2(2 + ey) = 3(x + 1)
