5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ integrate from here $\endgroup$ – Vasya Mar 27 '18 at 15:12
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Nikola Mar 27 '18 at 16:09
  • $\begingroup$ @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. $\endgroup$ – Jaideep Khare Mar 27 '18 at 16:55
3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Doesn't $d(xy) = y dx + xdy$ and not $xdx + y dy$? $\endgroup$ – Trevor Norton Mar 27 '18 at 15:17
  • 1
    $\begingroup$ 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. $\endgroup$ – Trevor Norton Mar 27 '18 at 15:30
  • $\begingroup$ @TrevorNorton You are right ... $\endgroup$ – Isham Mar 27 '18 at 15:46
  • $\begingroup$ @ÁngelMarioGallegos So we put all the expressions under the sign of the differential and than we integrate? A lot quicker, thanks. $\endgroup$ – Nikola Mar 27 '18 at 16:13
  • $\begingroup$ @Nikola, yes, it is a method known as integrating combinations. $\endgroup$ – Ángel Mario Gallegos Mar 27 '18 at 16:47
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.