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Easy differential equation Watch

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    The question asks to solve the ODE  \displaystyle \frac{dy}{dx}=\frac{1}{x^2-1} given that  y(0)=0 .
    Is the solution  \displaystyle \frac{1}{2} \ln \left | \frac{x-1}{x+1} \right | which would be valid for  x\neq \pm 1 OR
    Is the solution just  \displaystyle \frac{1}{2} \ln \frac{x-1}{x+1} which is valid for |x|<1?
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    Anyone?
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    (Original post by Ano123)
    The question asks to solve the ODE  \displaystyle \frac{dy}{dx}=\frac{1}{x^2-1} given that  y(0)=0 .
    Is the solution  \displaystyle \frac{1}{2} \ln \left | \frac{x-1}{x+1} \right | which would be valid for  x\neq \pm 1 OR
    Is the solution just  \displaystyle \frac{1}{2} \ln \frac{x-1}{x+1} which is valid for |x|<1?
    \int( \frac{1}{x^2 - 1})dx = -\int(\frac{1}{1-x^2})dx= -artanh(x) +c

    So the solution is negative of what you have and it is valid for  |x|&lt;1
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    (Original post by NotNotBatman)
    \int( \frac{1}{x^2 - 1})dx = -\int(\frac{1}{1-x^2})dx= -artanh(x) +c

    So the solution is negative of what you have and it is valid for  |x|&lt;1
    So in logarithmic form do we still need the absolute value in the log?
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    (Original post by Ano123)
    So in logarithmic form do we still need the absolute value in the log?
    No. \displaystyle -\frac{1}{2} \ln (\frac{1+x}{1-x})   = -\frac{1}{2}[ln(1+x) - ln(1-x)] Hopefully you can see that any of x substituted such that  -1&lt;x&lt;1 or |x|&lt;1 will have a positive argument in the logs, so there is no need for the absolute value.
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    (Original post by NotNotBatman)
    No. \displaystyle -\frac{1}{2} \ln (\frac{x-1}{x+1})   = -\frac{1}{2}[ln(x-1) - ln(x+1)] Hopefully you can see that any of x substituted such that  -1&lt;x&lt;1 or |x|&lt;1 will have a positive argument in the logs, so there is no need for the absolute value.
    x=1/2 gives a negative argument in the log.
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    (Original post by Ano123)
    x=1/2 gives a negative argument in the log.
    Typo, now corrected.
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    (Original post by NotNotBatman)
    Typo, now corrected.
    Does  \displaystyle \frac{1}{2} \ln \left \frac{1-x}{x+1} satisfy the ODE for |x|>1? Is this not a valid solution?
 
 
 
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