Solve the differential equation :
` (x + y) (dy)/(dx) = 1` .

Explanation

Given , ` (x + y) (dy)/(dx) = 1`
` therefore (dy)/(dx) = x + y`
`rArr (dx)/(dy) - x = y `
This is linear differential equation with x as
dependent veriable of the from ` (dx)/(dy) + P x = Q . `
` therefore P = - 1, Q = y `
The integrating factor ,
I.F. ` = e^(int" P dy" )= e^(int-1 dy) = e^(-y)`
` therefore ` The solution is ,
` x (I.F.) = int Q (I.F.) dy + C `
` rArr xe^(-y) = intye^(-y) dy + C `
` rArr xe^(-y) = y [-e^(-y)] - int 1 (-e^(-y)) dy + C `
`rArr ex^(-y) = - ye^(-y) - e^(-y) + C `
` rArr e^(-y) (x + y + 1) = C `
` rArr x + y + 1 = Ce^(y)` is general solution .