How to solve the non-linear ODE

Question

The ODE is $$\frac{d}{dx}\left(y(x)y'(x)\right)=-2xy'(x)$$

Answer 1

I doubt there is analytical solution for this. The following is a series solution. This is not hard to do since zero is an ordinary point. This assumes expansion around $x=0$

\begin{gather*} \boxed{{y^{\prime}}^{2}+y y^{\prime \prime}+2 y^{\prime} x=0} \end{gather*} With the expansion point at $x = 0$. Let

\begin{align*} y'' &= f\left( x,y,y^{\prime}\right)\\\\ &=-\frac{\left(y^{\prime}+2 x \right) y^{\prime}}{y} \end{align*}

Assuming expansion is at $x_{0}=0$ . Let initial conditions be $y\left( x_{0}\right) =y_{0}$ and $y^{\prime}\left( x_{0}\right) =y_{0}^{\prime}$. Using Taylor series gives

\begin{align*} y\left( x\right) & =y\left( x_{0}\right) +\left( x-x_{0}\right) y^{\prime}\left( x_{0}\right) +\frac{\left( x-x_{0}\right) ^{2}} {2}y^{\prime\prime}\left( x_{0}\right) +\frac{\left( x-x_{0}\right) ^{3} }{3!}y^{\prime\prime\prime}\left( x_{0}\right) +\cdots\\ & =y_{0}+xy_{0}^{\prime}+\frac{x^{2}}{2}\left. f\right\vert _{x_{0} ,y_{0},y_{0}^{\prime}}+\frac{x^{3}}{3!}\left. f^{\prime}\right\vert _{x_{0},y_{0},y_{0}^{\prime}}+\cdots\\ & =y_{0}+xy_{0}^{\prime}+\sum_{n=0}^{\infty}\frac{x^{n+2}}{\left( n+2\right) !}\left. \frac{d^{n}f}{dx^{n}}\right\vert _{x_{0},y_{0} ,y_{0}^{\prime}} \end{align*}

But \begin{align} \frac{df}{dx} & =\frac{\partial f}{\partial x}\frac{dx}{dx}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial y^{\prime}} \frac{dy^{\prime}}{dx}\tag{1}\\\\ & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y^{\prime }+\frac{\partial f}{\partial y^{\prime}}y^{\prime\prime}\\\\ & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y^{\prime }+\frac{\partial f}{\partial y^{\prime}}f\\\\ \frac{d^{2}f}{dx^{2}} & =\frac{d}{dx}\left( \frac{df}{dx}\right) \nonumber\\\\ & =\frac{\partial}{\partial x}\left( \frac{df}{dx}\right) +\frac{\partial }{\partial y}\left( \frac{df}{dx}\right) y^{\prime}+\frac{\partial}{\partial y^{\prime}}\left( \frac{df}{dx}\right) f\tag{2}\\\\ \frac{d^{3}f}{dx^{3}} & =\frac{d}{dx}\left( \frac{d^{2}f}{dx^{2}}\right) \nonumber\\\\ & =\frac{\partial}{\partial x}\left( \frac{d^{2}f}{dx^{2}}\right) +\left( \frac{\partial}{\partial y}\frac{d^{2}f}{dx^{2}}\right) y^{\prime} +\frac{\partial}{\partial y^{\prime}}\left( \frac{d^{2}f}{dx^{2}}\right) f\tag{3}\\\\ & \vdots\nonumber \end{align}

And so on. Hence if we name $F_{0}=f\left( x,y,y^{\prime}\right) $ then the above can be written as \begin{align} F_{0} & =f\left( x,y,y^{\prime}\right) \tag{4}\\\\ F_{1} & =\frac{df}{dx}\nonumber\\\\ & =\frac{dF_{0}}{dx}\nonumber\\\\ & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y^{\prime }+\frac{\partial f}{\partial y^{\prime}}y^{\prime\prime}\nonumber\\\\ & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y^{\prime }+\frac{\partial f}{\partial y^{\prime}}f\tag{5}\\\\ & =\frac{\partial F_{0}}{\partial x}+\frac{\partial F_{0}}{\partial y}y^{\prime}+\frac{\partial F_{0}}{\partial y^{\prime}}F_{0}\nonumber\\\\ F_{2} & =\frac{d}{dx}\left( \frac{d}{dx}f\right) \nonumber\\\\ & =\frac{d}{dx}\left( F_{1}\right) \nonumber\\\\ & =\frac{\partial}{\partial x}F_{1}+\left( \frac{\partial F_{1}}{\partial y}\right) y^{\prime}+\left( \frac{\partial F_{1}}{\partial y^{\prime} }\right) y^{\prime\prime}\nonumber\\\\ & =\frac{\partial}{\partial x}F_{1}+\left( \frac{\partial F_{1}}{\partial y}\right) y^{\prime}+\left( \frac{\partial F_{1}}{\partial y^{\prime} }\right) F_{0}\nonumber\\\\ & \vdots\nonumber\\\\ F_{n} & =\frac{d}{dx}\left( F_{n-1}\right) \nonumber\\\\ & =\frac{\partial}{\partial x}F_{n-1}+\left( \frac{\partial F_{n-1} }{\partial y}\right) y^{\prime}+\left( \frac{\partial F_{n-1}}{\partial y^{\prime}}\right) y^{\prime\prime}\nonumber\\\\ & =\frac{\partial}{\partial x}F_{n-1}+\left( \frac{\partial F_{n-1} }{\partial y}\right) y^{\prime}+\left( \frac{\partial F_{n-1}}{\partial y^{\prime}}\right) F_{0} \tag{6} \end{align}

Therefore 6) can be used from now on along with \begin{align*} y\left( x\right) &=y_{0}+xy_{0}^{\prime}+\sum_{n=0}^{\infty}\frac{x^{n+2} }{\left( n+2\right) !}\left. F_{n}\right\vert _{x_{0},y_{0},y_{0}^{\prime}} \tag{7} \end{align*} To find $y\left( x\right) $ series solution around $x=0$. Hence \begin{align*} F_0 &= -\frac{\left(y^{\prime}+2 x \right) y^{\prime}}{y}\\\\ F_1 &= \frac{d F_0}{dx} \\\\ &= \frac{\partial F_{0}}{\partial x}+ \frac{\partial F_{0}}{\partial y} y^{\prime}+ \frac{\partial F_{0}}{\partial y^{\prime}} F_0 \\\\ &= \frac{y^{\prime} \left(3 {y^{\prime}}^{2}+8 y^{\prime} x +4 x^{2}-2 y\right)}{y^{2}}\\\\ F_2 &= \frac{d F_1}{dx} \\\\ &= \frac{\partial F_{1}}{\partial x}+ \frac{\partial F_{1}}{\partial y} y^{\prime}+ \frac{\partial F_{1}}{\partial y^{\prime}} F_1 \\\\ &= -\frac{y^{\prime} \left(15 {y^{\prime}}^{3}+50 {y^{\prime}}^{2} x +44 x^{2} y^{\prime}+8 x^{3}-12 y^{\prime} y-12 y x \right)}{y^{3}}\\\\ F_3 &= \frac{d F_2}{dx} \\\\ &= \frac{\partial F_{2}}{\partial x}+ \frac{\partial F_{2}}{\partial y} y^{\prime}+ \frac{\partial F_{2}}{\partial y^{\prime}} F_2 \\\\ &= \frac{y^{\prime} \left(105 {y^{\prime}}^{4}+420 {y^{\prime}}^{3} x +520 {y^{\prime}}^{2} x^{2}+208 x^{3} y^{\prime}+16 x^{4}-98 {y^{\prime}}^{2} y-172 y^{\prime} y x -48 x^{2} y+12 y^{2}\right)}{y^{4}} \end{align*} And so on. Evaluating all the above at initial conditions $x = 0$ and $y \left(0\right) = y \left(0\right)$ and $y^{\prime}\left(0\right) = y^{\prime}\left(0\right)$ gives \begin{align*} F_0 &= -\frac{{y^{\prime}\left(0\right)}^{2}}{y \left(0\right)}\\\\ F_1 &= \frac{3 {y^{\prime}\left(0\right)}^{3}-2 y^{\prime}\left(0\right) y \left(0\right)}{y \left(0\right)^{2}}\\\\ F_2 &= \frac{-15 {y^{\prime}\left(0\right)}^{4}+12 {y^{\prime}\left(0\right)}^{2} y \left(0\right)}{y \left(0\right)^{3}}\\\\ F_3 &= \frac{105 {y^{\prime}\left(0\right)}^{5}-98 {y^{\prime}\left(0\right)}^{3} y \left(0\right)+12 y^{\prime}\left(0\right) y \left(0\right)^{2}}{y \left(0\right)^{4}} \end{align*} Substituting all the above in (7) and simplifying gives the solution as \begin{align*} y &= y \left(0\right) +x y^{\prime}\left(0\right)+ \frac{-\frac{{y^{\prime}\left(0\right)}^{2} x^{2}}{2}-\frac{x^{3} y^{\prime}\left(0\right)}{3}}{y \left(0\right)} +\frac{\frac{x^{3} {y^{\prime}\left(0\right)}^{3}}{2}+\frac{x^{4} {y^{\prime}\left(0\right)}^{2}}{2}+\frac{x^{5} y^{\prime}\left(0\right)}{10}}{y \left(0\right)^{2}}+\dots \end{align*}