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Math Help - i m stuck ... can anyone help me out to solve this ?

  1. #1
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    Post i m stuck ... can anyone help me out to solve this ?

    integrate the following
    y'=-2xy/(y2-x2)
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  2. #2
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    Re: i m stuck ... can anyone help me out to solve this ?

    What do you mean "integrate ..."? Are you trying to solve the differential equation? Let y = xv. Then you have:

    \dfrac{d(xv)}{dx} = \dfrac{2x(xv)}{x^2-(xv)^2} (Note: I factored a -1 from the denominator)

    v + x\dfrac{dv}{dx} = \dfrac{2v}{1-v^2}

    Separating variables:

    \dfrac{(1-v^2)dv}{v(v^2+1)} = \dfrac{dx}{x}

    Now, you can integrate both sides. On the left, use partial fractions: \dfrac{1-v^2}{v(v^2+1)} = \dfrac{1}{v} - \dfrac{2v}{v^2+1}.

    Once you solve for v(x), plug that in for y = xv. (Hint: to solve for v, you will need to use the quadratic formula).
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    Re: i m stuck ... can anyone help me out to solve this ?

    thankyou so much SlipEternal !! I got it..but can u explain me in what other conditions do we have to suppose y=xv or so ?
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  4. #4
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    Re: i m stuck ... can anyone help me out to solve this ?

    It is a common substitution to try. I am not sure if there is a specific "way to tell". I think practice is your best bet. The more you practice, the more likely you will notice a pattern among problems where that specific substitution works. Perhaps someone else on the forum has a better suggestion.
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    Re: i m stuck ... can anyone help me out to solve this ?

    im stuck again i couldn't solve for 'v'. I am getting complex numbers in result....can you please help me with this.I look forward on hearing from you soon.
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  6. #6
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    Re: i m stuck ... can anyone help me out to solve this ?

    So, you have:

    \int \dfrac{dv}{v} - \int \dfrac{2vdv}{v^2+1} = \int \dfrac{dx}{x}

    For the second integral, let u = v^2+1, du = 2vdv. Then you have:

    \int \dfrac{dv}{v} - \int \dfrac{du}{u} = \int \dfrac{dx}{x}

    \log(v) - \log(u) = \log(x) + C

    Plugging back in for u and using properties of log, you get:

    \log \left( \dfrac{v}{v^2+1} \right) = \log(x) + C

    Exponentiating both sides, you get:

    \dfrac{v}{v^2+1} = xe^C = kx where k = e^C.

    Multiplying out, you get:

    0 = kxv^2 - v + kx

    So, by the quadratic formula: v = \dfrac{1 \pm \sqrt{1-4k^2x^2}}{2kx}

    (If you are having trouble with coefficients to plug into the quadratic formula, 0 = av^2 + bv + c, you have a = kx, b = -1, c = kx)

    Plugging that in, you get: y = xv = \dfrac{1 \pm \sqrt{1-4k^2x^2}}{2k} which is defined over the reals for all -\dfrac{1}{2k} \le x \le \dfrac{1}{2k}.
    Last edited by SlipEternal; February 1st 2014 at 11:53 AM.
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    Re: i m stuck ... can anyone help me out to solve this ?

    got it thankyou so much!!
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    Re: i m stuck ... can anyone help me out to solve this ?

    You may want to check it out to make sure that it works:

    (2ky-1)^2 = 1-4k^2x^2

    Taking the derivative, you get:

    4k(2ky-1)y' = -8k^2x

    Solving for y', you get y' = \dfrac{-2kx}{2ky-1}, which is not equal to what you started with, so it seems likely that I made a mistake in my calculations somewhere.
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  9. #9
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    Re: i m stuck ... can anyone help me out to solve this ?

    haha yeah i didin't noticed that.. i will solve it again . anyways thanks for your help
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  10. #10
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    Re: i m stuck ... can anyone help me out to solve this ?

    Let's try that again:

    y = \dfrac{1 \pm \sqrt{1-4k^2x^2}}{2k}

    Let's simplify the constant: let K = \dfrac{1}{2k}. Then

    y = K \pm \dfrac{ \sqrt{(4k^2)\left(K^2 - x^2\right) } }{2k} = K \pm \sqrt{K^2-x^2}

    (That simplifies things greatly).

    Now, y' = \dfrac{\mp x}{\sqrt{K^2-x^2}}

    Let's see what we get from \dfrac{-2xy}{y^2-x^2}:

    \begin{align*}\dfrac{-2xy}{y^2-x^2} & = \dfrac{-2x(K \pm \sqrt{K^2-x^2})}{(K^2 \pm 2K\sqrt{K^2-x^2} + K^2-x^2) - x^2} \\ & = \dfrac{-x(K \pm \sqrt{K^2-x^2})}{\sqrt{K^2-x^2}(\sqrt{K^2-x^2} \pm K)} \\ & = \dfrac{\mp x}{\sqrt{K^2-x^2}} = y'\end{align*}

    So, it actually works out after all (just doesn't look like it will).
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