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

  1. #1
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    (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 ? (x+y ) dx +93x-3y-1) dy = 0-0061.jpg

    P/ s: the question is (x+y ) dx +(3x-3y-1) dy = 0
    Last edited by xl5899; Feb 15th 2016 at 10:08 PM.
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    Re: (x+y ) dx +93x-3y-1) dy = 0

    Hey xl5899.

    Can you show your work for the decomposition of your fraction of (2v-4)/(3v-4) please?
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    Re: (x+y ) dx +93x-3y-1) dy = 0

    Quote Originally Posted by chiro View Post
    Hey xl5899.

    Can you show your work for the decomposition of your fraction of (2v-4)/(3v-4) please?
    why ? i integrate (3v-4) / (2v-4) below in the working
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    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.
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    Re: (x+y ) dx +93x-3y-1) dy = 0

    Quote Originally Posted by chiro View Post
    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 ....
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    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.
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    Re: (x+y ) dx +93x-3y-1) dy = 0

    Quote Originally Posted by Archie View Post
    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 ???
    Last edited by xl5899; Feb 16th 2016 at 04:20 AM.
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