Answer 1

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

∴ `dy/dx = 1/(x+y)`

∴ `dy/dx = (x+y)`

∴ `dx/dy - x = y`

The given equation is of the form `dx/dy + Px = Q`

where, P = -1 and Q = y

∴ `I.F. = e int ^(pdy) = e int ^(-1dy) = e^-y`

∴ Solution of the given equation is

`x (I.F.) = int Q (I.F.) dy + c`

∴ `x e^-y = int ye^-y dy + c`

∴ `xe ^-y = y int e^-y dy - int [ d/dy(y)int e^-ydy] dy + c`

∴ `xe^-y =- y(e ^-y) - int 1xx(-e^-y) dy + c`

∴ `xe^-y = - ye ^-y -e ^-y +c`

∴  x = -y -1 + c e^y

∴  x + y + 1 = c e^y