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    I've been asked to integrate/"find general solution" of
    (x(y^(2) + 1)) *((x-1)y)^(-1)


    I'm struggling a lot with these last few chapters... Chapter 7 Differential Equations and 8 Curves defined Implicitly. Many of my class mates are stuggling too but that's little consolation :/ I don't really know how to start... was barely able to do Chapter 7 last week.
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    (Original post by one-final-hit)
    I've been asked to integrate/"find general solution" of
    (x(y^(2) + 1)) *((x-1)y)^(-1)


    I'm struggling a lot with these last few chapters... Chapter 7 Differential Equations and 8 Curves defined Implicitly. Many of my class mates are stuggling too but that's little consolation :/ I don't really know how to start... was barely able to do Chapter 7 last week.
    Is that

    \frac{dy}{dx}=\frac{x(y^2+1)}{y(  x-1)} ?

    If so, have you tried to separate the variables?
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    \large\frac {dy}{dx} = \frac {x(y^2+1)}{(x-1)y}
    \large\frac {y}{y^2+1} dy = \frac {x}{x-1} dx

    \large\int \frac {y}{y^2+1} dy = \int \frac {x}{x-1} dx
    \large\frac {1}{2} \int \frac {2y}{y^2+1} dy = \int \frac {x}{x-1} dx



    How about now ?
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    Yeh That was helpful; thank you to both of you.
    I've come to this now:
    1/2 * Ln(y^(2)+1) = x - 1 - ln(x-1) + C

    Not sure whether the right hand side is correct, and now what to do :/

    If the right hand side is correct, I've taken it down to:
    ln(y^(2)+1) = 2(x-1)-ln((x-1)^(2)) +c
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    (Original post by one-final-hit)
    Yeh That was helpful; thank you to both of you.
    I've come to this now:
    1/2 * Ln(y^(2)+1) = x - 1 - ln(x-1) + C

    Not sure whether the right hand side is correct, and now what to do :/

    If the right hand side is correct, I've taken it down to:
    ln(y^(2)+1) = 2(x-1)-ln((x-1)^(2)) +c
    Use integration by parts on the right hand side, and you should get (x-1)(ln(x-1)) instead

    The solution should be ln(y^(2)+1) = 2(x-1)(ln(x-1)) + C
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    (Original post by jameswhughes)
    Use integration by parts on the right hand side, and you should get (x-1)(ln(x-1)) instead

    The solution should be ln(y^(2)+1) = 2(x-1)(ln(x-1)) + C
    Can you explain how you used integration by parts because IMO that's the slower route I chose not to take simply because there was too much maths involved to justify it... im stuck either differentiating (x-1)^1 or integrating Ln(x-1).

    I figured out my mistake with u substitution in that I put x = u-1 instead of u+1, but my answer is still different to yours..
    = x-1+ln(x-1) + c

    Close to giving up seeing as I'm the only person in my class to have got this far...

    [edit: usubs]

    S x(x-1)^(-1) dx Where x-1=u, x=u+1, dx=1*du S= integral sign
    S ((u+1)/u) * 1 du
    S (u/u)+(1/u) du
    S 1+(1/u) du
    = [u +lnu +c]
    =x-1+ln(x-1) +c
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    (Original post by one-final-hit)
    Can you explain how you used integration by parts because IMO that's the slower route I chose not to take simply because there was too much maths involved to justify it... im stuck either differentiating (x-1)^1 or integrating Ln(x-1).

    I figured out my mistake with u substitution in that I put x = u-1 instead of u+1, but my answer is still different to yours..
    = x-1+ln(x-1) + c

    Close to giving up seeing as I'm the only person in my class to have got this far...
    Oops, looks as if I made a mistake :mad: Yes, your answer is right, just checked it
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    Ah ok, no problems
    mmm looking at the answer in the back of the book to try cook up a methd:
    y=+-sqrt((k(x-1)^(2) * e^(2x)) - 1)

    Can anyone tell me where that e^(2x) is coming from?
    [It's appeared once before in Chapter 7 Misc 7 Q8, and I didn't know what to do; looked in the book and it wasnt in that chapter so left it empty; teacher afaik, didn't explain it]
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    (Original post by one-final-hit)
    Ah ok, no problems
    mmm looking at the answer in the back of the book to try cook up a methd:
    y=+-sqrt((k(x-1)^(2) * e^(2x)) - 1)

    Can anyone tell me where that e^(2x) is coming from?
    [It's appeared once before in Chapter 7 Misc 7 Q8, and I didn't know what to do; looked in the book and it wasnt in that chapter so left it empty; teacher afaik, didn't explain it]
    I think they have just multiplied by e, as [k1]e^(2x-1)=[k2]e^2x if the k2 is just another constant.
 
 
 
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