# Thread: (x+y ) dx +93x-3y-1) dy = 0

1. ## (x+y ) dx +93x-3y-1) dy = 0

I am asked to solve this question using transformation of v = x + y .

the ans given is x +3y + 2 ln (y+x -2) , but my ans is x +3y + 2 ln (2y+2x -4) , anyone can point out where's my mistake ?

P/ s: the question is (x+y ) dx +(3x-3y-1) dy = 0

Hey xl5899.

3. ## Re: (x+y ) dx +93x-3y-1) dy = 0

Originally Posted by chiro
Hey xl5899.

why ? i integrate (3v-4) / (2v-4) below in the working

4. ## Re: (x+y ) dx +93x-3y-1) dy = 0

I want to see what you get after you adjust it for integration.

The integral should involve integrated (3v-4)/(2v-4) which has a partial fraction decomposition.

This working will determine if there is an error in the integral being calculated.

5. ## Re: (x+y ) dx +93x-3y-1) dy = 0

Originally Posted by chiro
I want to see what you get after you adjust it for integration.

The integral should involve integrated (3v-4)/(2v-4) which has a partial fraction decomposition.

This working will determine if there is an error in the integral being calculated.
(3v-4)/(2v-4) = 1.5 +2 / (2v-4) , i already show it in my working ....

6. ## Re: (x+y ) dx +93x-3y-1) dy = 0

Where you get $\ln |2v-4|$ you could write $\ln |2v-4|-\ln2$, where we have just subtracted a part of the arbitrary constant of integration. Alternatively, you can cancel the 2 before integrating.

7. ## Re: (x+y ) dx +93x-3y-1) dy = 0

Originally Posted by Archie
Where you get $\ln |2v-4|$ you could write $\ln |2v-4|-\ln2$, where we have just subtracted a part of the arbitrary constant of integration. Alternatively, you can cancel the 2 before integrating.
if i do nit cancel off the 2 before , is it okay to leave the ans as (x+3y) +2ln(2x+2y-4) = C , instead of (x+3y) +2ln(x+y-2) = C ???