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

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

integrate the following
y'=-2xy/(y2-x2)

2. ## 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).

3. ## 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 ?

4. ## 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.

5. ## 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.

6. ## 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 $-\dfrac{1}{2k} \le x \le \dfrac{1}{2k}$.

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

got it thankyou so much!!

8. ## 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.

9. ## 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

10. ## 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).