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Math Help - another question!

  1. #1
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    another question!

    I am having problem solving this problem. Kind of stuck. Help would be veryy appreciated.
    the problem is :
    The differential equation y' = x^2 - y^2 -1 has y = -x as one of its solutions find the another solution.
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  2. #2
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    Quote Originally Posted by blazingwes View Post
    I am having problem solving this problem. Kind of stuck. Help would be veryy appreciated.
    the problem is :
    The differential equation y' = x^2 - y^2 -1 has y = -x as one of its solutions find the another solution.
    Your DE has the general form \frac{dy}{dx} = p(x) + q(x) y + r(x) y^2 (known as a Riccati equation).

    The usual technique for this type of DE is to make the substitution z = \frac{1}{y - f(x)} where y = f(x) is a particular solution to the DE.
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  3. #3
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    i used one of the solution and
    i let y = u - x
    therefore y' = u'-1

    plug it back into the equation and i got
    u'-1 = x^2 - ( u-x)^2 -1
    u'-1 = x^2 - (u^2-2ux+x^2)-1
    u'-1 = x^2 - u^2 + 2ux - x^2 -1
    u' = -u^2+2ux
    u' = u(-u+2x)

    I dunno if i am doing it correctly or not... i am stuck at the integration part
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  4. #4
    Super Member Random Variable's Avatar
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    I ran into a problem.


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

    Doing as mr fantastic suggested, let  z = \frac{1}{y+x}

    then  y = \frac {1}{z} - x

     \frac {dy}{dx} = \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1

     \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1 = x^{2} - \Big(\frac{1}{z} - x \Big)^{2} -1

     \frac {\text{-}1}{z^{2}} \frac {dz}{dx} = \frac {\text{-}1}{z^{2}} + \frac {2x}{z}

     \frac {dz}{dx} = 1 - 2xz

     \frac {dz}{dx} + 2xz = 1

    now we have a first order linear differential equation

     e^{x^{2}} \frac {dz}{dx} + 2xz e^{x^{2}} = e^{x^{2}}

     \frac {d}{dx} e^{x^{2}}z = e^{x^{2}} \ dx

     e^{x^{2}}z = \int e^{x^{2}} \ dx ?
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  5. #5
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    continuing from what i did.... i got the same answer too
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  6. #6
    Super Member Random Variable's Avatar
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    The most you can do is leave the answer as  z = e^{\text{-}x^{2}} \Big(\int_{x_{0}}^{x} e^{t^{2}} \ dt + C \Big)

    then  \frac {1}{y+x} = e^{\text{-}x^{2}} \Big(\int_{x_{0}}^{x} e^{t^{2}} \ dt + C\Big)

    or  y = \frac {1}{e^{\text{-}x^{2}} \Big(\int_{x_{0}}^{x} e^{t^{2}} \ dt + C\Big)}- x

    yuck
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  7. #7
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    lol yahhh yuck
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  8. #8
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    Quote Originally Posted by blazingwes View Post
    lol yahhh yuck
    - Wolfram|Alpha
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  9. #9
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    Quote Originally Posted by Random Variable View Post
    I ran into a problem.


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

    Doing as mr fantastic suggested, let  z = \frac{1}{y+x}

    then  y = \frac {1}{z} - x

     \frac {dy}{dx} = \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1

     \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1 = x^{2} - \Big(\frac{1}{z} - x \Big)^{2} -1

     \frac {\text{-}1}{z^{2}} \frac {dz}{dx} = \frac {\text{-}1}{z^{2}} + \frac {2x}{z}

     \frac {dz}{dx} = 1 - 2xz

     \frac {dz}{dx} + 2xz = 1

    now we have a first order linear differential equation

     e^{x^{2}} \frac {dz}{dx} + 2xz e^{x^{2}} = e^{x^{2}}

     \frac {d}{dx} e^{x^{2}}z = e^{x^{2}} \ dx

     e^{x^{2}}z = \int e^{x^{2}} \ dx ?
    and then use:

    \text{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{t^2}\ dt

    to simplify this.

    CB
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