# 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

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

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

$\dfrac{dy}{dx}=\dfrac{-xy\pm\sqrt{x^2y^2+4xy-4x}}{2}$

Let $u=xy$ ,

Then $y=\dfrac{u}{x}$

$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$

$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\dfrac{-u\pm\sqrt{u^2+4u-4x}}{2}$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{\pm\sqrt{u^2+4u-4x}-u}{2}+\dfrac{u}{x^2}$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{(\pm\sqrt{u^2+4u-4x}-u)x^2+2u}{2x^2}$

$((\pm\sqrt{u^2+4u-4x}-u)x^2+2u)\dfrac{dx}{du}=2x$

Let $v=\pm\sqrt{u^2+4u-4x}$ ,

Then $x=\dfrac{u^2}{4}+u-\dfrac{v^2}{4}$

$\dfrac{dx}{du}=\dfrac{u}{2}+1-\dfrac{v}{2}\dfrac{dv}{du}$

$\therefore\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)\left(\dfrac{u}{2}+1-\dfrac{v}{2}\dfrac{dv}{du}\right)=2\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)$

$\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)\left(\dfrac{u}{2}+1\right)-\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)\dfrac{v}{2}\dfrac{dv}{du}=\dfrac{u^2}{2}+2u-\dfrac{v^2}{2}$

$\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)\dfrac{v}{2}\dfrac{dv}{du}=\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)\left(\dfrac{u}{2}+1\right)+\dfrac{v^2}{2}-\dfrac{u^2}{2}-2u$

$\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)v\dfrac{dv}{du}=\left((v-u)\left(\dfrac{u^2}{4}+u-\dfrac{v^2}{4}\right)^2+2u\right)(u+2)+v^2-u^2-4u$

And follow the approach in Can anyone transform this ODE into a different form?.

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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 $$y = 1-\frac{1}{2}\,{x}^{2}+\frac{1}{6}\,{x}^{3}+{\frac {7}{48}}\,{x}^{4}+{\frac {1}{240}} \,{x}^{5}+{\frac {1}{2160}}\,{x}^{6}+{\frac {787}{30240}}\,{x}^{7}+{ \frac {20047}{645120}}\,{x}^{8}+{\frac {370693}{10450944}}\,{x}^{9} + \ldots$$
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