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

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.

• integrate from here – Vasya Mar 27 '18 at 15:12

## 5 Answers

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

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

• and from there I can exponentiate the equation. What stopped me from doing this was the arbitrary constant. Is this a sufficient solution, i.e. having an unknown constant? – Nikola Mar 27 '18 at 16:09
• @Nikola Yes, it is sufficient. Actually this constant is just similar to constant of integration. Infact, we have integrated "0" , which therefore results in a constant. – Jaideep Khare Mar 27 '18 at 16:55

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

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 $$\boxed{x^2+y^2+y^2x^2=K}$$

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

• Doesn't $d(xy) = y dx + xdy$ and not $xdx + y dy$? – Trevor Norton Mar 27 '18 at 15:17
• That's what I'm saying. If we take the differential, we get $d(xy + \frac 1 2 x^2 y^2) = y dx + xdy + xy^2 dx + yx^2 dy$, which is not the same as the original $xdx + ydy + xy^2dx + yx^2dy=0$. I think the answer should be $x^2y^2 + x^2 + y^2 = c$ instead. – Trevor Norton Mar 27 '18 at 15:30
• @TrevorNorton You are right ... – Isham Mar 27 '18 at 15:46
• @ÁngelMarioGallegos So we put all the expressions under the sign of the differential and than we integrate? A lot quicker, thanks. – Nikola Mar 27 '18 at 16:13
• @Nikola, yes, it is a method known as integrating combinations. – Ángel Mario Gallegos Mar 27 '18 at 16:47

$$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$ $$(1+x^2)y\,dy =-(1+y^2)x\,dx$$ 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.

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