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Math Help - Simple first order ordinary differential equation

  1. #1
    Senior Member chella182's Avatar
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    Simple first order ordinary differential equation

    Okay, so this question should be easy but for some reason I'm having major bother with it.

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

    I got some sort of answer which I don't even think is right, and what I got I just couldn't simplify at all I know it's separable, hence why I should be able to do it, but I just can't.
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  2. #2
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    I think the problem is the integration of \frac1{1-x^2}. Any ideas how to tackle it?
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  3. #3
    Senior Member chella182's Avatar
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    Well it's difference of 2 squares, and I did that, then partial fractions and what not, but I really don't think my answer was right.
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  4. #4
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    No partial fractions, just algebra: \frac{1}{1-x^{2}}=\frac{(1+x)+(1-x)}{2(1+x)(1-x)}=\frac{1}{2}\left( \frac{1}{1-x}+\frac{1}{1+x} \right).
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chella182 View Post
    Okay, so this question should be easy but for some reason I'm having major bother with it.

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

    I got some sort of answer which I don't even think is right, and what I got I just couldn't simplify at all I know it's separable, hence why I should be able to do it, but I just can't.
    Yes it is separable...but I have a feeling we can't find an explicit solution (i.e. y=...)

    The DE is equivalent to \frac{\,dy}{y^2-1}=\frac{\,dx}{x^2-1}\implies \frac{\,dy}{(y+1)(y-1)}=\frac{\,dx}{(x+1)(x-1)}

    Now, we see that if we integrate, we have \int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}.

    Note that (y+1)-(y-1)=2 and (x+1)-(x-1)=2

    Therefore, \int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}\implies\tfrac{1}{2}\int\frac{2\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{2\,dx}{(x+1)(x-1)} \implies\tfrac{1}{2}\int\frac{\left[(y+1)-(y-1)\right]\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{\left[(x+1)-(x-1)\right]\,dx}{(x+1)(x-1)}.

    This now becomes \int\frac{\,dy}{y-1}-\int\frac{\,dy}{y+1}=\int\frac{\,dx}{x-1}-\int\frac{\,dx}{x+1} \implies\ln\left|y-1\right|-\ln\left|y+1\right|=\ln\left|x-1\right|-\ln\left|x+1\right|+C\implies \ln\left|\frac{y-1}{y+1}\right|=\ln\left|\frac{x-1}{x+1}\right|+C

    This format is fine...

    Otherwise, you can say this is the same as

    \left|\frac{y-1}{y+1}\right|=C\left|\frac{x-1}{x+1}\right|\implies\frac{y-1}{y+1}=A\left|\frac{x-1}{x+1}\right| \implies \left[1-A\left|\frac{x-1}{x+1}\right|\right]y=A\left|\frac{x-1}{x+1}\right|+1\implies y=\frac{A\left|\displaystyle\frac{x-1}{x+1}\right|+1}{1-A\left|\displaystyle\frac{x-1}{x+1}\right|} where A=\pm C

    I hope this helps...
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  6. #6
    Senior Member chella182's Avatar
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    Partial fractions is algebra isn't it? Anyway that's what I got, but then integrating both sides and whatever went messy & I just couldn't do it.

    Quote Originally Posted by Chris L T521 View Post
    Yes it is separable...but I have a feeling we can't find an explicit solution (i.e. y=...)

    The DE is equivalent to \frac{\,dy}{y^2-1}=\frac{\,dx}{x^2-1}\implies \frac{\,dy}{(y+1)(y-1)}=\frac{\,dx}{(x+1)(x-1)}

    Now, we see that if we integrate, we have \int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}.

    Note that (y+1)-(y-1)=2 and (x+1)-(x-1)=2

    Therefore, \int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}\implies\tfrac{1}{2}\int\frac{2\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{2\,dx}{(x+1)(x-1)} \implies\tfrac{1}{2}\int\frac{\left[(y+1)-(y-1)\right]\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{\left[(x+1)-(x-1)\right]\,dx}{(x+1)(x-1)}.

    This now becomes \int\frac{\,dy}{y-1}-\int\frac{\,dy}{y+1}=\int\frac{\,dx}{x-1}-\int\frac{\,dx}{x+1} \implies\ln\left|y-1\right|-\ln\left|y+1\right|=\ln\left|x-1\right|-\ln\left|x+1\right|+C\implies \ln\left|\frac{y-1}{y+1}\right|=\ln\left|\frac{x-1}{x+1}\right|+C

    This format is fine...

    Otherwise, you can say this is the same as

    \left|\frac{y-1}{y+1}\right|=C\left|\frac{x-1}{x+1}\right|\implies\frac{y-1}{y+1}=A\left|\frac{x-1}{x+1}\right| \implies \left[1-A\left|\frac{x-1}{x+1}\right|\right]y=A\left|\frac{x-1}{x+1}\right|+1\implies y=\frac{A\left|\displaystyle\frac{x-1}{x+1}\right|+1}{1-A\left|\displaystyle\frac{x-1}{x+1}\right|} where A=\pm C

    I hope this helps...

    Totally don't get that bit in the middle about things equalling 2. That's not how we've been taught to do stuff like that :\. I did get halves kicking about, but argh! Think I'm gonna have to go to the lecturer :\.
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  7. #7
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    Quote Originally Posted by chella182 View Post
    Partial fractions is algebra isn't it? Anyway that's what I got, but then integrating both sides and whatever went messy & I just couldn't do it.
    Chella - you're on the right track - it's partial fractions!

    \frac{A}{x-1} + \frac{B}{x+1} = \frac{1}{(x-1)(x+1)}

    A(x+1) + B(x-1)= 1

    (A+B)x + (A-B) = 1\;\;\;\Rightarrow\;\;\;A+B=0,\;\;A-B=1

    so

     A = \frac{1}{2},\;\;\; b= - \frac{1}{2}

    and

    \frac{1}{2} \frac{1}{x-1} - \frac{1}{2} \frac{1}{x+1}

    Lot's of students don't see " that bit in the middle about things equalling 2" but the method above works fairly well!
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  8. #8
    Senior Member chella182's Avatar
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    They're both \frac{1}{2} aren't they because it's 1-x in one of the brackets, not x-1
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  9. #9
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    Quote Originally Posted by chella182 View Post
    They're both \frac{1}{2} aren't they because it's 1-x in one of the brackets, not x-1
    It depends what you want to split up

    \frac{1}{(1-x)(1+x)} or \frac{1}{(x-1)(x+1)}

    the signs will change accordingly.
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  10. #10
    Math Engineering Student
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    It's

    Quote Originally Posted by chella182 View Post

    (1-x^2)\frac{dy}{dx}=1-y^2
    Quote Originally Posted by Krizalid View Post
    No partial fractions, just algebra: \frac{1}{1-x^{2}}=\frac{(1+x)+(1-x)}{2(1+x)(1-x)}=\frac{1}{2}\left( \frac{1}{1-x}+\frac{1}{1+x} \right).
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  11. #11
    Senior Member chella182's Avatar
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    Someone had a minus in between them up there I'm sure. Urgh, I'm leaving it for tonight.
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