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This is what I have, but I am not sure if I am on right track, or if more work needs to be done:

Rearranging the given D.E. I have:

$ v = \frac{1}{x^2} = \frac{x}{2x-x'}$

so:

$ v' = -\frac{2}{x^3} = -\frac{2}{2x-x'}$

Substituting for x', I have:

$v' = \frac{2}{x^3} = \frac{2v}{x}$

Is this sufficient? Or do I need to elminate x from the equation completely? Is there more work to be done?

Thanks in advance!!!

3 Answers 3

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You really need to eliminate $x$ completely. To that end, you need to express $v$ and $v'$ in terms of $x$.

This is the shortest path I'm thinking about right now.

You have

$x'+x^3=2x$

Multiply by $x$, to get

$xx'+x^4=2x^2$

But since $x^2=v^{-1}$,

$xx'+v^{-2}=2v^{-1}$

And finally

$2xx'=-v^{-2}v'$

so

$2xx'+2v^{-2}=4v^{-1}$

$-v^{-2}v'+2v^{-2}=4v^{-1}$

So mulipling by $v^2$ gives

$-v'+2=4v$

  • 0
    I understand that, but could you show me what the intermediate step is between $ xx' + v^{-2} = 2v^{-1}$ and $2xx'=-v^{-2}v'$? Sorry, its just not getting through to me.2012-10-13
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Differentiate with respect to $t$ so you get $2xx^{\prime} = -v^{-2}v^{\prime}$

  • 0
    Although there is nothing wrong with your answer per se, it is rather sparse, especially when compared to the other answer from a year ago. That said, I like sparse answers! However, they are better when they appear as answers to new questions (as who would read a sparse one when they have a complete solution to read?). You can find lots of unanswered questions [here](http://math.stackexchange.com/unanswered).2013-10-14
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At points such that $x\ne0$, $x'+x^3=2x$ is equivalent to $\dfrac{x'}{x^3}+1=\dfrac2{x^2}$ which is equivalent to $-\dfrac12\left(\dfrac1{x^2}\right)'+1=\dfrac2{x^2}$ which reads $-\dfrac12v'+1=2v$.