# First order ordinary differential equations involving powers of the slope

Are there any general approaches to differential equations like

$$x-x\ y(x)+y'(x)\ (y'(x)+x\ y(x))=0,$$

or that equation specifically?

The problem seems to be the term $y'(x)^2$. Solving the equation for $y'(x)$ like a qudratic equation gives some expression $y'(x)=F(y(x),x)$, where $F$ is "not too bad" as it involves small polynomials in $x$ and $y$ and roots of such object. That might be a starting point for a numerical approach, but I'm actually more interested in theory now.

$y(x)=1$ is a stationary solution. Plugging in $y(x)\equiv 1+z(x)$ and taking a look at the new equation makes me think functions of the form $\exp{(a\ x^n)}$ might be involved, but that's only speculation. I see no symmetry whatsoever and dimensional analysis fails.

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I am not a specialist in ODEs, but I remember that since brute force approaches usually imply expanding in power series, the squares in $y$ or $y'$ (or the product of these two) are usually considered very annoying. The equations that are considered "nice enough" are those which after a nice change of variable are linear with respect to the $y$'s and $y'$'s. The problem with your solution (which I have already experimented) is that you don't get a general solution as easily as you would think. – Patrick Da Silva Jun 25 '12 at 21:56
I'm actually more interested in theory now -- but what do you expect to get from theory? Stability of solutions around $y=1$? Their global existence or blow-up? Asymptotic expansion as $x\to\infty$? – user31373 Jun 25 '12 at 23:14

Hint:

$x-xy+y'(y'+xy)=0$

$(y')^2+xyy'+x-xy=0$

$(y')^2=x(y-1-yy')$

$x=\dfrac{(y')^2}{y-1-yy'}$

Let $t=y'$ ,

Then $\left(1+\dfrac{t^3(t-1)}{y^2(t-1)+1}\right)\dfrac{dy}{dt}=\dfrac{2t^2}{y(1-t)-1}+\dfrac{yt^3}{y(1-t^2)-1}$

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The usual existence-and-uniqueness theory has problems because $y'$ is not a Lipschitz function of $x$ and $y$. In particular, besides $y(x)=1$ the initial value $y(0)=1$ seems to have a series solution of the form $$y = 1-\frac{1}{2}\,{x}^{2}+\frac{1}{6}\,{x}^{3}+{\frac {7}{48}}\,{x}^{4}+{\frac {1}{240}} \,{x}^{5}+{\frac {1}{2160}}\,{x}^{6}+{\frac {787}{30240}}\,{x}^{7}+{ \frac {20047}{645120}}\,{x}^{8}+{\frac {370693}{10450944}}\,{x}^{9} + \ldots$$

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It turns out that this solution and $y(x)=1$ belong to different choices of $\pm$ in $y' = (- x y \pm \sqrt{x^2 y^2 + 4 x y - 4 x})/2$: $y=1$ for $+$ when $x > 0$ and $-$ when $x < 0$, the solution above for $-$ when $x > 0$ and $+$ when $x < 0$. – Robert Israel Jun 27 '12 at 2:25