How find y from $x^2 y^3 d x+x\left(1+y^2\right) d y=0$

Question

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:

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

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

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

4. Why D must be connected set?

Thank you in advance.

Answer 1

$$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$ 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 : $$y(x)=\pm\frac{1}{\sqrt{W\left(C\:e^{x^2} \right)}}=\pm\frac{1}{\sqrt{W\left(\:e^{x^2+2C'} \right)}}$$ $C=e^{2C'}$