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Thread: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

  1. #1
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    (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    can someone check thru the working ? I couldn't figure out which part i did wrongly . (x^2)(y+1)dx + (y^2)(x-1)dy = 0-0900.jpg
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    Two minus signs dropped when you rationalise the fractions in line three.
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    Quote Originally Posted by Archie View Post
    Two minus signs dropped when you rationalise the fractions in line three.
    i still cant find the mistake ? what is the correct one ?
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    Quote Originally Posted by Archie View Post
    Two minus signs dropped when you rationalise the fractions in line three.
    well , i have redo the question , yet i still cant get (x+1)^2 and (y-1)^2 .......(x^2)(y+1)dx + (y^2)(x-1)dy = 0-0062.jpg
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    You just need to add 2 two both sides. Then the right hand side $c+2 = c_1$ is still an arbitrary constant.
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    Quote Originally Posted by Archie View Post
    You just need to add 2 two both sides. Then the right hand side $c+2 = c_1$ is still an arbitrary constant.


    why the author wanna do so ?
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    So it looks nicer. Or perhaps he integrated $\displaystyle (x-1)$ as a single term. You could even do substitutions before you start.
    Last edited by Archie; Feb 17th 2016 at 02:26 AM.
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    Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    Quote Originally Posted by Archie View Post
    So it looks nicer. Or perhaps he integrated $\displaystyle (x-1)$ as a single term. You could even do substitutions before you start.
    can you explain further?
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