Let's continue a bit with your solution. So
and exponentiating gives
.
Your IC of gives and your solution is
(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!!!
The answer in the book says that the general solution is
x+3y+2 ln(2-x-y)=c.. the LN is taken of the whole expression in the parentheses. Where did the x and 3y go in your solution?
So we're solving for c, not for y.
The end result should be x+3y+2ln(2-x-y)=equals some value.
Thanks for your help!