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I have derived this equation, but it doesn't seem to be possible to solve it analytically, I don't see a way to separate $y$ and $x$, unless there is another way I am not aware of in this instance. Moreover, even when I try to represent it as a slope field on the specified domain I get nothing. Does it mean that the solution does not exist? Is there, maybe, a theorem or some other way to prove it more rigorously?

$$2(p-y(x))y(x)+\frac{x(x-2p)+y(x)^2}{y'(x)}=0$$

where $p$ - a positive parameter on $[0,1]$ and $x \in [p,1]$, $y \in [0,1]$.

  • 0
    Where are the derivatives in this equation? – 2017-02-28
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    $y'(x)$ in the denominator of the fraction? – 2017-02-28
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    In fact $y(x)=x$ is the solution to this equation, but it doesn't make sense, because this equation is ought to describe something descending... – 2017-02-28
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    You can rewrite $y(x)^2(1-2y'(x)) + x^2-2px+2py(x)y'(x) = 0$, maybe it makes better sense. – 2017-02-28
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    Hm... I am not sure how does it help? – 2017-02-28

1 Answers 1

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Hint:

$2(p-y(x))y(x)+\dfrac{x(x-2p)+y(x)^2}{y'(x)}=0$

$2(y-p)y\dfrac{dy}{dx}=x(x-2p)+y^2$

Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=166:

Let $u=\dfrac{y}{x}$ ,

Then $y=xu$

$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

$\therefore2(xu-p)xu\left(x\dfrac{du}{dx}+u\right)=x(x-2p)+(xu)^2$

$2(xu-p)x^2u\dfrac{du}{dx}+2(xu-p)xu^2=x(x-2p)+x^2u^2$

$2(xu-p)x^2u\dfrac{du}{dx}=x(x-2p)-(2xu-x-2p)xu^2$

$(x-2p-(2xu-x-2p)u^2)\dfrac{dx}{du}=2(xu-p)xu$

$((1-(2u-1)u^2)x+2p(u^2-1))\dfrac{dx}{du}=2u^2x^2-2pux$

This belongs to an Abel equation of the second kind.