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Idea of how to separate?

  1. Dec 10, 2006 #1
    Problem: [tex](y-yx^2)\frac{dy}{dx} = (y+1)^2[/tex]

    So, the first thing I tried was just dividing the whole equation by [tex](y-yx^2)[/tex] and then factored out the y to get [tex]\frac{dy}{dx} = \frac{(y+1)^2}{y(1-x^2)}[/tex]. Next I expanded the numerator on the right side of the equation and then split them all into idividual fractions to get:

    [tex]\frac{dy}{dx} = \frac{y}{1-x^2}+\frac{2}{1-x^2}+\frac{1}{y(1-x^2)}[/tex].

    I just don't see how I can go from here?
     
  2. jcsd
  3. Dec 10, 2006 #2
    Ok, so I worked it out some more and got [tex]\frac{ydy}{(y+1)^2} = \frac{dx}{(1-x^2)}[/tex]

    Is this correct?
     
  4. Dec 10, 2006 #3

    cristo

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    Yes, that looks right to me
     
  5. Dec 10, 2006 #4
    Cool Thanks. So I worked this one out and got the left hand side of the equation, but I can seem to get the integration on the right.

    [tex]\frac{ydy}{(y+1)^2} => \frac{dx}{(1-x^2)} = \int{\frac{ydy}{(y+1)^2}} = \int{\frac{dx}{(1-x^2)}}[/tex]

    so, [tex]ln|y+1|+\frac{1}{y+1} = \int{\frac{dx}{(1-x^2)}}[/tex]

    what can I do with this one?

    * For some reason on my screen the y integral keeps showing up in the post even though it is not in the LaTex. So, if you see it too, please disregard the middle term that reads [tex]\int{\frac{ydy}{(y+1)^2}}[/tex] in the last part of the problem. Thanks
     
    Last edited: Dec 10, 2006
  6. Dec 10, 2006 #5

    cristo

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    Thats not the correct integral of the LHS. Have you come across the method of partial fractions?
     
  7. Dec 10, 2006 #6
    Yes I have worked with partial fractions a little bit, but I don't see how it applies to this one. I don't mean to second guess you, but are you sure the LHS is not correct? I worked it out again and got the same answer and my TI-89 gives the same answer as well.
     
  8. Dec 10, 2006 #7

    cristo

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    I get a factor 1/2, but it might just be me. How did you integrate that by hand without using partial fraction, just out of curiosity?! EDIT: the factor 1/2 is just me!! sorry

    For the x integral, consider the denominator (1-x^2). This can be written as
    (1+x)(1-x). So, if you can express [tex] \frac{1}{1-x^2} [/tex] as a sum[tex] \frac{a}{1+x}+\frac{b}{1-x} [/tex] then this can be integrated directly
     
    Last edited: Dec 10, 2006
  9. Dec 10, 2006 #8
    Thanks, I will try that out. To get the LHS w/o partial fractions:

    [tex]\int{\frac{ydy}{(y+1)^2}}[/tex]

    let u = y+1, du = dy, and y = u-1

    So, [tex]\int{\frac{(u-1)}{(u)^2}du}[/tex]

    =

    [tex]\int{\frac{u}{u^2}-\frac{1}{u^2}du} [/tex]

    =

    [tex] ln|u| + \frac{1}{u}[/tex]

    Substitute y+1 for u and get [tex]ln|y+1|+\frac{1}{y+1}[/tex]
     
    Last edited: Dec 10, 2006
  10. Dec 10, 2006 #9

    cristo

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    Ahh, ok, nice. I didn't spot that! I would generally use partial fractions for expressions like that, but looks like the sub'n is simpler there!
     
  11. Dec 10, 2006 #10
    Oh man, haha.. I like the substitution method, but I am not too keen on the trig substitution. Although, this one seems pretty nice for that. Do you mind taking a look at my partial fraction work? I need some practice with it so I was wondering if I did it correctly.

    [tex]\int{\frac{dx}{(1-x^2)}}[/tex]

    [tex]\frac{1}{(1+x)(1-x)}=\frac{A}{1+x}+\frac{B}{1-x}[/tex]

    So, B and A both = (1/2)

    Then the solution to the integral is:
    [tex]\frac{ln|1+x|}{2}+\frac{ln|1-x|}{2}[/tex]

    Thanks!!
     
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