# Solve $x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$

I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.

Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$

Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$

Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$

This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?

Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$

-
Please, avoid subjective titles and look for more informative, objective, descriptive titles. –  Pedro Tamaroff Mar 14 '13 at 1:23

$x=y\dfrac{dy}{dx}-\left(\dfrac{dy}{dx}\right)^2$

$\left(\dfrac{dy}{dx}\right)^2-y\dfrac{dy}{dx}+x=0$

Apply the method in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=2:

Let $F(x,y,t)=t^2-yt+x~,$

Then $\dfrac{dy}{dt}=-\dfrac{t\dfrac{\partial F}{\partial t}}{\dfrac{\partial F}{\partial x}+t\dfrac{\partial F}{\partial y}}=-\dfrac{t(2t-y)}{1+t(-t)}=\dfrac{2t^2}{t^2-1}-\dfrac{ty}{t^2-1}$

$\dfrac{dy}{dt}+\dfrac{ty}{t^2-1}=\dfrac{2t^2}{t^2-1}$

$y=t+\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{\sqrt{t^2-1}}$

$\therefore\dfrac{dx}{dt}=-\dfrac{\dfrac{\partial F}{\partial t}}{\dfrac{\partial F}{\partial x}+t\dfrac{\partial F}{\partial y}}=-\dfrac{2t-y}{1+t(-t)}=\dfrac{2t}{t^2-1}-\dfrac{y}{t^2-1}=\dfrac{t}{t^2-1}-\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{(t^2-1)^\frac{3}{2}}$

$x=\int\biggl(\dfrac{t}{t^2-1}-\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{(t^2-1)^\frac{3}{2}}\biggr)dt=\dfrac{t(\ln(t+\sqrt{t^2-1})+C_1)}{\sqrt{t^2-1}}+C_2$

Hence $\begin{cases}x=\dfrac{t(\ln(t+\sqrt{t^2-1})+C_1)}{\sqrt{t^2-1}}+C_2\\y=t+\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{\sqrt{t^2-1}}\end{cases}$

-

Note that $$dx=\frac{dy}{p} \tag{*}$$ Since you have found $y$ in terms of $p$, you can find $dy$, substitute in $(*)$ and obtain the solution in a somewhat more palatable parametric form.

-
I've carried out the computation you suggested (see my answer) but I don't see how it helps. Could you please be a little more detailed? Unless you're saying we can write $y$ and $x$ parametrically in terms of $p$ - can we do that? –  Daniel Littlewood Mar 13 '13 at 21:04
yes, that is precisely how it works. I have to admit I have not carried out the complete calculation, but this is the general outcome of this method; generally the explicit solution will be rarely possible –  Valentin Mar 13 '13 at 21:13
Doesn't it matter that $p=\frac{dy}{dx}$ is not independent of $y$ or $x$? –  Daniel Littlewood Mar 13 '13 at 21:23
Check out this: en.wikipedia.org/wiki/Clairaut's_equation –  Valentin Mar 13 '13 at 21:34

complete squares , i think this equation can be written as

$$-x+ \frac{y^{2}}{4}= ( \frac{dy}{dx}- \frac{y}{2})^{2}$$

take square roots so in one side we are left with

$$\sqrt{ -x +\frac{{y}^{2}}{4}}$$

on the other side we have $$\frac{dy}{dx}- \frac{y}{2}$$

remember the square root has two solutin with +1 and -1 now the equation

-
Yes, but how would we solve this equation? –  Daniel Littlewood Mar 13 '13 at 20:17