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Thread: mixed fractured equations

  1. #1
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    mixed fractured equations

    a.
    $\displaystyle y'+\frac{2}{x}y=\frac{y^{3}}{x^{2}} $

    b.
    $\displaystyle 2y'-\frac{x}{y}=\frac{xy}{x^{2}-1}$
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  2. #2
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    $\displaystyle \displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{1}{x^2}\,y^3$.

    This is a Bernoulli Equation. So make the substitution $\displaystyle \displaystyle v = y^{-2}$.

    Then $\displaystyle \displaystyle y = v^{-\frac{1}{2}}$ and

    $\displaystyle \displaystyle \frac{dy}{dx} = \frac{d}{dx}(v^{-\frac{1}{2}})$

    $\displaystyle \displaystyle = \frac{d}{dv}(v^{-\frac{1}{2}})\,\frac{dv}{dx}$

    $\displaystyle \displaystyle = -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dx}$.


    Then the DE becomes

    $\displaystyle \displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{1}{x^2}\,y^3$

    $\displaystyle \displaystyle -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dx} + \frac{2}{x}\,v^{-\frac{1}{2}} = \frac{1}{x^2}\,v^{-\frac{3}{2}}$

    $\displaystyle \displaystyle \frac{dv}{dx} - \frac{4}{x}\,v = -\frac{2}{x^2}$.

    This is now first order linear, so use the integrating factor method.

    The integrating factor is

    $\displaystyle \displaystyle e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{(x^{-4})}} = x^{-4}$.


    Multiplying through by the integrating factor gives

    $\displaystyle \displaystyle x^{-4}\,\frac{dv}{dx} - 4x^{-5}\,v = -2x^{-6}$

    $\displaystyle \displaystyle \frac{d}{dx}(x^{-4}\,v) = -2x^{-6}$

    $\displaystyle \displaystyle x^{-4}\,v = \int{-2x^{-6}\,dx}$

    $\displaystyle \displaystyle x^{-4}\,v = \frac{2}{5}x^{-5}$

    $\displaystyle \displaystyle v = \frac{2}{5}x^{-1}$.


    Since $\displaystyle \displaystyle v = y^{-2}$

    $\displaystyle \displaystyle \frac{2}{5}x^{-1} = y^{-2}$

    $\displaystyle \displaystyle \left(\frac{2}{5}x^{-1}\right)^{-\frac{1}{2}} = y$

    $\displaystyle \displaystyle y = \left(\frac{5}{2}\right)^{\frac{1}{2}}x^{\frac{1}{ 2}}$

    $\displaystyle \displaystyle y = \frac{\sqrt{10x}}{2}$
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  3. #3
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    how to solve b?

    i tried to recombinate it into
    (-xy^2-x^3+x)dx+2y(x^2-1)dy=0

    and solve it this way

    is there any other easier way
    ?
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  4. #4
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    (b) is also Bernoulli. Multiply by $\displaystyle y$ and try letting $\displaystyle u = y^2.$
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  5. #5
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    in bernulyy method we eliminate y from the right side and if the power of y is 3 then we divide by y^3
    and put a new variable v=1/y^2

    so we need to divide by y

    but is our variable
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  6. #6
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    Quote Originally Posted by Danny View Post
    (b) is also Bernoulli. Multiply by $\displaystyle y$ and try letting $\displaystyle u = y^2.$
    How can it be Bernoulli? When you multiply by $\displaystyle \displaystyle y$ you get

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

    which is not of the correct form $\displaystyle \displaystyle \frac{dy}{dx} + P(x)\,y = Q(x)\,y^n$...
    Last edited by Prove It; Nov 13th 2010 at 11:32 PM.
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  7. #7
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    You sure? Look closely.
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  8. #8
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    Well I had a sign wrong, but apart from that, it does not look to be of the correct form...
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  9. #9
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    i am confused
    when i multiply by y i get a bernully for
    correct?
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    i am confused
    when i multiply by y i get a bernully for
    correct?
    As has been said already, you multiply it by y and then make the substitution u = y^2:

    $\displaystyle \displaystyle \frac{du}{dx} - x = \frac{x}{x^2 - 1} u$

    This is a simple first order ODE solvable using the integrating factor method.
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  11. #11
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    thanks
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