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Math Help - General solution of differential equation - question 2

  1. #1
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    General solution of differential equation - question 2

    I have the question

    2) Find the general solution of the following differential equation:

    >> dy/dx = y^2 -1/x

    i have fount the general solution to be:

    re arrange>> 1/y^2 -1 dy = -1/x dx
    integrate>> 1/2 ln(y-1) - 1/2 ln(y+1) = lnx + C

    is that correct?? Or can i simplify further??
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  2. #2
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    Re: General solution of differential equation - question 2

    Quote Originally Posted by NFS1 View Post
    I have the question

    2) Find the general solution of the following differential equation:

    >> dy/dx = y^2 -1/x

    i have fount the general solution to be:

    re arrange>> 1/y^2 -1 dy = -1/x dx
    integrate>> 1/2 ln(y-1) - 1/2 ln(y+1) = lnx + C

    is that correct?? Or can i simplify further??
    Dear NFS1,

    Your differential equation is not clear. Do you mean,

    \frac{dy}{dx}=\frac{y^2-1}{x}\mbox{ or }\frac{dy}{dx}=y^2-\frac{1}{x} ?
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  3. #3
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    Re: General solution of differential equation - question 2

    Quote Originally Posted by Sudharaka View Post
    Dear NFS1,

    Your differential equation is not clear. Do you mean,

    \frac{dy}{dx}=\frac{y^2-1}{x}\mbox{ or }\frac{dy}{dx}=y^2-\frac{1}{x} ?
    For the OP, if it's the first, then it's separable. If it's the second, it's a Bernoulli equation.
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    Re: General solution of differential equation - question 2

    Quote Originally Posted by Sudharaka View Post
    Dear NFS1,

    Your differential equation is not clear. Do you mean,

    \frac{dy}{dx}=\frac{y^2-1}{x}\mbox{ or }\frac{dy}{dx}=y^2-\frac{1}{x} ?
    Its the first one mate...i have seperated and got my final answer as...1/2 ln(y-1) - 1/2 ln(y+1) = lnx + C
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: General solution of differential equation - question 2

    What you wrote looks fine, but don't forget the modulus signs where necessary.
    Now you have to determine the general solution ...

    Can you do that? ...
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  6. #6
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    Re: General solution of differential equation - question 2

    what i wrote does it simplify to >> ln (y-1/y+1) = ln x + C ??
    i dont quite understand what to do from here??
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: General solution of differential equation - question 2

    Yes, that's correct and write C=\ln(e^C) (you don't need modulus signs here because e^C>0, so you get:
    \ln\left|\frac{y-1}{y+1}\right|=2\ln|x|+2\ln(e^C)
    \Leftrightarrow \ln\left|\frac{y-1}{y+1}\right|= \ln\left(x^2\cdot e^{2C}\right)
    \Leftrightarrow \left|\frac{y-1}{y+1}\right|=x^2\cdot e^{2C}
    \Leftrightarrow \left|1-\frac{2}{y+1}\right|=x^2\cdot e^{2C}
    \Leftrightarrow y=...

    Try to finish it ...
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  8. #8
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    Re: General solution of differential equation - question 2

    y = (-1/x^2.e^2C) - 1 ???
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    Re: General solution of differential equation - question 2

    I would say:
    1-\frac{2}{y+1}=\pm x^2 \cdot e^{2C}, let \pm e^{2C}=K
    \Leftrightarrow \frac{2}{y+1}=1-(x^2\cdot K)
    \Leftrightarrow y+1=\frac{2}{1-(x^2\cdot K)}
    \Leftrightarrow y=\frac{2}{1-(x^2\cdot K)}-1
    \Leftrigharrow y= \frac{2-(1-x^2\cdot K)}{1-x^2\cdot K}
    \Leftrightarrow y=\frac{1+x^2\cdot K}{1-x^2\cdot K}
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  10. #10
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    Re: General solution of differential equation - question 2

    Quote Originally Posted by Siron View Post
    I would say:
    1-\frac{2}{y+1}=\pm x^2 \cdot e^{2C}, let \pm e^{2C}=K
    \Leftrightarrow \frac{2}{y+1}=1-(x^2\cdot K)
    \Leftrightarrow y+1=\frac{2}{1-(x^2\cdot K)}
    \Leftrightarrow y=\frac{2}{1-(x^2\cdot K)}-1
    \Leftrigharrow y= \frac{2-(1-x^2\cdot K)}{1-x^2\cdot K}
    \Leftrightarrow y=\frac{1+x^2\cdot K}{1-x^2\cdot K}
    Just a quick question why does ... y-1/y+1 become 1-2/y+1 ??
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  11. #11
    MHF Contributor Siron's Avatar
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    Re: General solution of differential equation - question 2

    Because:
    1-\frac{2}{y+1}=\frac{y+1}{y+1}-\frac{2}{y+1}=\frac{y+1-2}{y+1}=\frac{y-1}{y+1}
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