<s>[INST] How find y from $x^2 y^3 d x+x\left(1+y^2\right) d y=0$ Suppose $\alpha (x,y)=\frac{1}{xy^3}$ is integral factor of equation $$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$
Check $\alpha (x,y)$:
$x^2 y^3 d x+x\left(1+y^2\right) d y=0 \mid \cdot \frac{1}{x y^3} \quad x \neq 0, y \neq 0$
$x d x+y^3\left(1+y^2\right) d y=0 \quad (\square)$
Let $M(x)=x, \quad \text{and} \quad N(x)=y^3\left(1+y^2\right)$ where $
D=\left\{ \left( x;y \right) \in \mathbb{R} ^2\,\,| \begin{matrix}
 x>0&  \land&  y>0\\
\end{matrix} \right\} 
$
I notice:

*

*$D$  is connected set

*$M(x),\: N(x)$ are continuous function over $D$ ($C^1$)

Let us check that the partial derivatives are equal:
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
therefore
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
Back to $ (\square)$:
$\int M(x, y) d x+\int N(x, y) d y=\mathbf{C}$
$\int x d x+\int \frac{1+y^2}{y^3} d y=\mathbf{C}$
$\frac{x^2}{2}+\int\left(\frac{1}{y^3}+\frac{1}{y}\right) d y=\mathbf{C}$
$\frac{x^2}{2}+\frac{y^{-3+1}}{-3+1}+\ln (y)=\mathbf{C}$
$\frac{x^2}{2}-\frac{1}{2 y^2}+\ln (y)=\mathbf{C}$
Question:

*

*How find solutions for the variable $y$ like WolframAlpha. I know WolframAlpha use Lambert function, but i don't know find y.
$$
y=-\frac{1}{\sqrt{W\left(e^{x^2-2 C}\right)}}
$$
$$
y=\frac{1}{\sqrt{W\left(e^{x^2-2 C}\right)}}
$$


*What happens if $(0,0) \in R^2$. Is solution differential equation?


*What happens if $ (x, y) \in R^2 - D$ without (0,0)?


*Why D must be connected set?
Thank you in advance.
 [/INST] $$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$
Separable ODE :
$$xdx=-\left(\frac{1}{y^3}+\frac{1}{y} \right)dy$$
Integrate :
$$\frac12 x^2=\frac{1}{2y^2}+\ln|y|+\text{constant}$$
Solution on the form of inverse function :
$$x(y)=\pm\sqrt{\frac{1}{y^2}+2\ln|y|+c}$$
Explicit solution :
$$\ln|y|=\frac12 x^2+\frac{1}{2y^2}+\text{constant}$$
$$y^2=C\:e^{x^2}e^{1/y^2}$$
$$y^{-2}e^{y^{-2}}=C\:e^{x^2}$$
From the definition of the LambertW function $Ae^A=B\implies A=W(B)$
with $A=y^{-2}$ and $B=C\:e^{x^2}$ we get :
$$y^{-2}=W\left( C\:e^{x^2}\right)$$
Finally the WolframAlpha solution :
$$y(x)=\pm\frac{1}{\sqrt{W\left(C\:e^{x^2} \right)}}=\pm\frac{1}{\sqrt{W\left(\:e^{x^2+2C'} \right)}}$$
$C=e^{2C'}$
</s>