solving $y' - yy'x^2-x=0$ How can i solve this?
$$y' - yy'x^2-x=0$$
I only got to the homogeneous solution wich I found is (I just divided by $y'$) 
$$y=\frac{1}{x^2}$$
But I don't know how to get the particular solution, I have for certain that it's not a constant as I tried to find it in every way possibile, could anybody help me? Thanks.
 A: $$y'-yy'x^{ 2 }-x=0\\ { y }^{ \prime  }\left( 1-y{ x }^{ 2 } \right) -x=0\\ \frac { dy }{ dx } \left( 1-y{ x }^{ 2 } \right) -x=0\\ 1-y{ x }^{ 2 }-x{ x }^{ \prime  }=0\\ z={ x }^{ 2 }\Rightarrow { z }^{ \prime  }=2x{ x }^{ \prime  }$$ $$1-yz-\frac { { z }^{ \prime  } }{ 2 } =0$$ 
which  becomes to the "Ordinary differential equation 
" respect to the  $z$

$$2-2yz-{ z }^{ \prime  }=0$$ 

$$2-2yz-{ z }^{ \prime  }=0\\ { z }^{ \prime  }=-2yz\\ \frac { { z }^{ \prime  } }{ z } =-2y\\ \int { \frac { dz }{ z } =-2\int { ydy }  } \\ \ln { \left| z \right|  } =-{ y }^{ 2 }+C\\ z=C{ e }^{ -{ y }^{ 2 } }\\ z=C(y){ e }^{ -{ y }^{ 2 } }\\ { z }^{ \prime  }={ C }^{ \prime  }{ (y)e }^{ -{ y }^{ 2 } }-2y{ e }^{ -{ y }^{ 2 } }C(y)\\ 2-2yC(y){ e }^{ -{ y }^{ 2 } }-{ C }^{ \prime  }{ (y)e }^{ -{ y }^{ 2 } }+2y{ e }^{ -{ y }^{ 2 } }C(y)=0\\ { C }^{ \prime  }{ (y)e }^{ -{ y }^{ 2 } }=2\\ { C }^{ \prime  }(y)=2e^{ { y }^{ 2 } }\\ C(y)=2\int { { e }^{ { y }^{ 2 } } } dy+C\\ z={ e }^{ -{ y }^{ 2 } }\left( 2\int { { e }^{ { y }^{ 2 } } } dy+C \right) $$
so the final answer is :

$$\\ { x }^{ 2 }={ e }^{ -{ y }^{ 2 } }\left( 2\int { { e }^{ { y }^{ 2 } } } dy+C \right) $$

A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\imp}{\Longrightarrow}
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 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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Write th equation $\ds{y' - yy'x^{2} - x = 0}$ as
$\ds{-x\,\dd x + \pars{1 - x^{2}y}\dd y = 0}$. Multiply both members by
$\ds{\mathrm{C}\pars{x,y}}$. It means
$\ds{-\mathrm{C}\pars{x,y}x\,\dd x + \mathrm{C}\pars{x,y}\pars{1 - x^{2}y}\dd y
     = 0}$ and set
\begin{align}
\partiald{\bracks{-\mathrm{C}\pars{x,y}x}}{y} & =
\partiald{\bracks{\mathrm{C}\pars{x,y}\pars{1 - x^{2}y}}}{x}
\\[3mm]
\imp\quad-\,\partiald{\mathrm{C}\pars{x,y}}{y}\,x & =
\partiald{\mathrm{C}\pars{x,y}}{x}\pars{1 - x^{2}y}-2xy\,\mathrm{C}\pars{x,y}
\end{align}
It is convenient, by simplicity, to choose $C$ as a function of $y$ ( independent of $x$ ) such that $\mathrm{C}\pars{y} = \expo{y^{2}}$. It means that
$$
\dd\Phi = -x\expo{y^{2}}\,\dd x + \expo{y^{2}}\pars{1 - x^{2}y}\dd y = 0
\quad\mbox{is an exact differential.}
$$
Then,
\begin{align}
\partiald{\Phi}{x} & = -x\expo{y^{2}}\quad\imp\quad\Phi = -\,\half\,x^{2}\expo{y^{2}} + \mathrm{f}\pars{y}
\\[3mm]
\partiald{\Phi}{y} & = -x^{2}y\expo{y^{2}} + \mathrm{f}'\pars{y} = \expo{y^{2}}\pars{1 - x^{2}y}\quad\imp\quad\mathrm{f}'\pars{y} = \expo{y^{2}}\ \imp\
\mathrm{f}\pars{y} = \int\expo{y^{2}}\,\dd y
\end{align}
So, your solution is given implicitily by
$$
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad\mbox{constant} = \Phi =
-\,\half\,x^{2}\expo{y^{2}} + \int\expo{y^{2}}\,\dd y\quad}
\\ \mbox{} \\ \hline
\end{array}
$$
A: An approach if you are OK with a power series representation is to pause at this step:
$$y'(1-yx^2) = x \\\text{or}\\ y'-x^2y'y=x$$
Now assume $$y = \sum_{k=-\infty}^\infty c_kx^k$$
Subtraction, differentiation w.r.t. $x$ and multiplication with $x^2$ behaves nicely on the coefficients in the space of linear combination of monomials. The tricky/confusing part will be the product $y'(1-yx^2)$. It will turn into a (shifted) self-convolution (with inverse sign) for our coefficients. The key here will be that right hand side is very simple. Only one 1-value ($1\cdot x$) and all other are 0.
For almost all positions there will be an equation one particular coefficient equals negative scalar product of two permuted coefficient vectors. Gosh this got more confusing then I thought. I think I will need to ask a question on my own for the resolution of this.
