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Thread: 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?).

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

    Using partial fractions, you should have

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

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

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

    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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