What is the integrating factor of $ydx-(y^{2}+x^{2}+x)dy=0$

Question

I study ODEs from Tenenbaum and Pollard textbook and come to this problem in exercise 10 on integrating factors .

$ydx-(y^{2}+x^{2}+x)dy=0$

I tried to check the condition of these integrating factors

$h(x) \: ,h(y) \: ,h(xy) \: ,h(x/y) \: ,h(y/x)$ , but all failed

Answer 1

$$ydx-(y^{2}+x^{2}+x)dy=0 \tag 1$$ An integrating factor is $\frac{1}{x^2+y^2}$

$\frac{1}{x^2+y^2}ydx-\frac{1}{x^2+y^2}(y^{2}+x^{2}+x)dy=dF \tag 2$

$\frac{\partial F}{\partial x}=\frac{1}{x^2+y^2}y \quad\to\quad \frac{\partial^2 F}{\partial y\partial x}=\frac{x^2-y^2}{(x^2+y^2)^2}$

$\frac{\partial F}{\partial y}=-\frac{1}{x^2+y^2}(y^{2}+x^{2}+x)\quad\to\quad \frac{\partial^2 F}{\partial x\partial y}=\frac{x^2-y^2}{(x^2+y^2)^2}$

$\frac{\partial^2 F}{\partial y\partial x}=\frac{\partial^2 F}{\partial x\partial y}\quad$ thus Eq.$(2)$ is a total derivative.

$F=\int \frac{y}{x^2+y^2}dx=\tan^{-1}\left(\frac{x}{y}\right)+f(y)=\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right)+f(y)$

$F=\int \frac{-(y^2+x^2+x)}{x^2+y^2}dy=-y-\tan^{-1}\left(\frac{y}{x}\right)+g(x)$

This implies : $\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right)+f(y)=-y-\tan^{-1}\left(\frac{y}{x}\right)+g(x)$

$g(x)=0$ and $f(y)=-y-\frac{\pi}{2} \quad\to\quad F=y-\tan^{-1}\left(\frac{y}{x}\right)=c$

The solution of EDO $(1)$ , expressed on the form of implicite equation is : $$y-\tan^{-1}\left(\frac{y}{x}\right)=c$$ Or, explicitly as $x$ function of $y$ : $$x=\frac{y}{\tan(y-c)} $$

Answer 2

write your equation in the form $$\frac {1}{\frac{dy}{dx}}=\frac{y}{y^2+x(y)^2+x(y)}$$ and then $$\frac{dx(y)}{dy}=y+\frac{x(y)^2}{y}+\frac{x(y)}{y}$$ let $$x(y)=yv(y)$$ then we have $$\frac{\frac{dv(y)}{dy}}{v(y)^2+1}=1$$ can you finish?

Answer 3

This is not an answer to the question, but a comment too long to be put in the comments section.

The answer to the question about integrating factor was already given and so, the solution of the ODE can be derived. But it is not the easier way to solve the ODE. I would like show a simpler method.

$$ydx-(y^{2}+x^{2}+x)dy=0$$ is in fact a Riccati's ODE : $$\frac{dx}{dy}=\frac{1}{y}x^{2}+\frac{1}{y}x+y$$ In order to transform it to a linear equation, the usual change of variable is : $x(y)=\frac{u'(y)}{-\frac{1}{y}u(y)}= -y\frac{u'}{u}$

$\frac{dx}{dy}=-\frac{u'}{u}-y\frac{u''}{u}+y\frac{(u')^2}{u^2} = \frac{1}{y}\left(-y\frac{u'}{u}\right)^{2}+\frac{1}{y}\left(-y\frac{u'}{u}\right)+y$

After simplification : $\quad\frac{u''}{u} = -1 \quad\to\quad u=c_1\cos(y)+c_2\sin(y)$ $$x(y)= -y\frac{-c_1\sin(y)+c_2\cos(y)}{c_1\cos(y)+c_2\sin(y)}=y\tan(y+C)$$ with $\quad\tan(C)=-\frac{c_2}{c_1}$.

The solution of the initial ODE on the form of implicit equation is : $$x-y\tan(y+C)=0$$