Results 1 to 10 of 10
Like Tree1Thanks
  • 1 Post By SlipEternal

Thread: i m stuck ... can anyone help me out to solve this ?

  1. #1
    Newbie
    Joined
    Jul 2013
    From
    kathmandu
    Posts
    15

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

    integrate the following
    y'=-2xy/(y2-x2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,397
    Thanks
    1351

    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 $\displaystyle y = xv$. Then you have:

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

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

    Separating variables:

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

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

    Once you solve for $\displaystyle v(x)$, plug that in for $\displaystyle y = xv$. (Hint: to solve for v, you will need to use the quadratic formula).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2013
    From
    kathmandu
    Posts
    15

    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 ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,397
    Thanks
    1351

    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.
    Thanks from baidyasuyam
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2013
    From
    kathmandu
    Posts
    15

    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,397
    Thanks
    1351

    Re: i m stuck ... can anyone help me out to solve this ?

    So, you have:

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

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

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

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

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

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

    Exponentiating both sides, you get:

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

    Multiplying out, you get:

    $\displaystyle 0 = kxv^2 - v + kx$

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

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

    Plugging that in, you get: $\displaystyle y = xv = \dfrac{1 \pm \sqrt{1-4k^2x^2}}{2k}$ which is defined over the reals for all $\displaystyle -\dfrac{1}{2k} \le x \le \dfrac{1}{2k}$.
    Last edited by SlipEternal; Feb 1st 2014 at 11:53 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2013
    From
    kathmandu
    Posts
    15

    Re: i m stuck ... can anyone help me out to solve this ?

    got it thankyou so much!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,397
    Thanks
    1351

    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:

    $\displaystyle (2ky-1)^2 = 1-4k^2x^2$

    Taking the derivative, you get:

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

    Solving for y', you get $\displaystyle y' = \dfrac{-2kx}{2ky-1}$, which is not equal to what you started with, so it seems likely that I made a mistake in my calculations somewhere.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jul 2013
    From
    kathmandu
    Posts
    15

    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
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,397
    Thanks
    1351

    Re: i m stuck ... can anyone help me out to solve this ?

    Let's try that again:

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

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

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

    (That simplifies things greatly).

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

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

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

    So, it actually works out after all (just doesn't look like it will).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Sep 26th 2012, 07:41 AM
  2. Trying to solve by substitution...getting stuck near the end
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Jun 30th 2012, 04:44 PM
  3. Replies: 1
    Last Post: Jan 16th 2012, 04:10 AM
  4. Stuck on 'solve the equation...'
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 27th 2007, 06:30 AM
  5. Stuck, Stuck, Stuck - Need Help Urgently
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 17th 2005, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum