Find a Particular Solution, given initial condition

**(x+y)dx+(3x+3y-4)dy=o ---> y(1)=1**

x+y=u

dy=du-dx

3u-4=3x+3y-4

udx+ (3u-u)(du-dx)=0

Solving, we arrive at

dx + (3u-4)/(4-2u)=0

x- 1/2(2 ln(u-2)+ 3u)=c

2x-(2 ln (u-2) + 3u)=c

Substituting

2x-2ln(x+y-2) - 3 (x+y)=c

-x-3y-2ln(x+y-2)=c

x+3y+2ln(x+y-2)=c

__So here are the problems I ran into__:

The answer in the book has the following answer:

x+3y+2ln(2-x-y)=c

If you plug in the Initial Value for x and y, you get

1+3+ 2Ln(2-1-1)= 4+2Ln(0), which obviously creates a problem since LN(0) if undefined.

Help Please!!!