Given:
dy/dx= (x+y+3)/(2x+2y+1)
Let v = x+y
∴ y = v-x
Put the equation in terms of v and x by eliminating y:
d(v-x)/dx = (v + 3)/(2x + 2(v-x) + 1)
(dv/dx) - 1 = (v + 3)/(2x + 2v - 2x + 1)
Combine like terms:
(dv/dx) - 1 = (v + 3)/(2v + 1)
dv/dx = (v + 3)/(2v + 1) + 1
dv/dx = (3v + 4)/(2v + 1)
∫[(2v + 1)/(3v + 4)]dv = ∫dx
Splitting (2v + 1)/(3v + 4) into two terms using technique for partial fraction:
(2v + 1)/(3v + 4) = 2/3 - (5/3)(3v + 4)
Therefore:
(2/3)∫dv - (5/3)∫dv/(3v + 4)= ∫dx
2v/3 - (5/9)•ln(3v + 4) = x + C
Since v = x + y
(2x+2y)/3 - (5/9)•ln(3x+3y +4) = x + C
