Question

 Solve the differential equation:

`("x" + "y") "dy"/"dx" = 1`

Answer 1

`("x" + "y") "dy"/"dx" = 1`

`"dy"/"dx" = 1/("x" +"y") => "dx"/"dy" = "x" + "y"`

`"dx"/"dy" + (-1) "x" = "y"`

Comparing with `"dx"/"dy" + "P"."x" = "Q" "we get"`

P = -1 and Q = y

`therefore = "e"^(int-1"dy") = "e"^-"y"`

The general solution is 

x (IF) = ∫ Q (IF) dy +C

xe^-y = y ∫ e^-y dy +C

⇒ xe^-y = -ye^-y - e^-y + C

⇒ (x + y + 1) = ce^y