solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$

Question

I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$

Dividing by $dx$ we have

$x + xy^2 + yy' + yy'x^2=0$

From where,

$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\frac{1}{2} d\ln(1+x^2)=\frac{1}{2}d\ln(1+y^2)(1+x^2)=0$$

Let $c=(1+y^2)(1+x^2)$, so our equation becomes: $$ d\ln c=0 $$

So what should I do here, should I integrate, or should I divide by $dx$?

If I divide by dx I get the expression $2x+2yy'+2xy^2+2x^2yy'=0$ which has $x$, $y$ and $y'$ and doesn't help me get anywhere.

Thanks in advance.

Answer 1

$$d( \text{something})=0 \implies \text{something = constant}$$

So you get the solution $$\ln(1+y^2)(1+x^2) = C$$ (Where $C$ is arbitrary constant)

Answer 2

The DE is $$\frac12d(x^2)+\frac12d(y^2)+\frac12d(x^2y^2)=0$$ Then, the solution is $$\boxed{\frac12x^2+\frac12y^2+\frac12x^2y^2=c}$$

Answer 3

it is $$-\frac{y'(x)}{\frac{1+y(x)^2}{y(x)}}=\frac{x}{1+x^2}$$

Answer 4

$$d\ln c=0 \implies \ln(c)=K$$

Another Hint

$$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$ $$(x + xy^2)dx + (y + yx^2)dy=0$$ It's an exact differential... $$\frac {\partial P}{\partial y}=\frac {\partial Q}{\partial x} \implies 2xy=2xy$$ $$ \begin{cases} f(x,y)=\int x+xy^2dx \\ f(x,y)=\int y+yx^2dy \end{cases} $$ Therefore $$\boxed{x^2+y^2+y^2x^2=K}$$

Answer 5

$$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$ $$(1+x^2)y\,dy =-(1+y^2)x\,dx $$ If $x^2,y^2\ne-1$, above equation can be written as (after multiplying both side by $2$): $$\frac{2y}{1+y^2}\,dy =-\frac{2x}{1+x^2}\,dx $$ $$\int \frac{2y}{1+y^2}\,dy =-\int \frac{2x}{1+x^2}\,dx $$ $$\ln (1+y^2)=-\ln (1+x^2)+\ln c=\ln \frac {c}{1+x^2}$$ where $c$ is a constant. Hence, $$1+y^2= \frac {c}{1+x^2}$$ $$y^2= \frac {c}{1+x^2}-1$$ Or: $$y^2+x^2+y^2x^2= C$$ where $C$ is also a constant.