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Math Help - simplification/possible error

  1. #1
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    simplification/possible error

    Hi,

    I have the differential equation (1-x^2)y' = 1 - y^2.
    Using partial fractions and exponentials, i now have the equation in the following form:

    sqrt(1+y)/sqrt(1-y) = A(sqrt(1+x)/sqrt(1-x))

    I am not sure if i have simplified this enough, as the square roots are causing me problems.
    Also i am not sure if I have used the right method. I have used seperation, but I'm not sure if i could have used a homogeneous equation method.

    Any thoughts or comments would be appreciated.
    Thanks

    stephen
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by schteve View Post
    Hi,

    I have the differential equation (1-x^2)y' = 1 - y^2.
    Using partial fractions and exponentials, i now have the equation in the following form:

    sqrt(1+y)/sqrt(1-y) = A(sqrt(1+x)/sqrt(1-x))

    I am not sure if i have simplified this enough, as the square roots are causing me problems.
    Also i am not sure if I have used the right method. I have used seperation, but I'm not sure if i could have used a homogeneous equation method.

    Any thoughts or comments would be appreciated.
    Thanks

    stephen
    Separation is the way to go about this, since this equation is not homogeneous (do you know why?).

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

    Using partial fractions, you should have

    \frac{1}{2}\left[\frac1{1+y}+\frac1{1-y}\right]\,dy=\frac{1}{2}\left[\frac1{1+x}+\frac1{1-x}\right]\,dx \implies \ln|1+y|-\ln|1-y|=\ln|1+x|-\ln|1-x|+C\implies \left|\frac{1+y}{1-y}\right|=C\left|\frac{1+x}{1-x}\right|

    At this stage, let's get rid of absolute values on the LHS:

    \frac{1+y}{1-y}=\pm C\left|\frac{1+x}{1-x}\right|\implies 1+y=C\left|\frac{1+x}{1-x}\right|-C\left|\frac{1+x}{1-x}\right|y \implies \left(C\left|\frac{1+x}{1-x}\right|+1\right)y=C\left|\frac{1+x}{1-x}\right|-1\implies y=\frac{C\left|\dfrac{1+x}{1-x}\right|-1}{C\left|\dfrac{1+x}{1-x}\right|+1}

    This looks quite complicated, but at least we have a value for y. Does this part make sense?
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  3. #3
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    Hi

    Thanks, I think it's not homogeneous because all the terms are not of the same degree - i think that's right?

    Also, I forgot the halves can cancel, hence why i have sqaure roots in my example.
    Another thought has also occured to me, and that is that you could convert the integrals into arctan expressions? I don't know which is the best way to approach the integral now.

    Again thanks fo the help
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