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!!!

Let's continue a bit with your solution. So


and exponentiating gives

.

Your IC of gives and your solution is

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!

The two answers





are equivalent however yours cannot recover the initial condition where as mine can.

This problem is similar to finding the solutions to

subject to (i) , and (ii) .