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I'd love your help with solving this following differential equation: $(2x^3y^2-y)dx+(2x^2y^3-x)dy=0.$

I tried to use check if this is an exact equation and find a integration, but it didn't work. Then I tried to divide the equation by some factor and to use some kind of assignment such as $z=\frac{y}{x}$, but it didn't work for me either.

Any suggestions?

($y$ is a function of $x$)

Even though Julian answer is great, I wonder if there's a solution without an integration depends both on x and y.

Thanks a lot!

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    Solutions that I've found $y=x$, $x=0$2012-04-08

3 Answers 3

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$\mu(x,y)=\frac{1}{x^2 y^2}$ is an integrating factor. The solution is $x^2+y^2+ \frac{1}{xy}=c$.

I found the integrating factor using the Prelle-Singer algorithm.

I write $\dot x = x-2x^2 y^3 =f_1 $

$\dot y=2x^3 y^2-y=f_2$

It is clear that $x$ and $y$ are Darboux polynomials with cofactor $\Lambda_1=2x y^3-1$ and $\Lambda_2=2 y x^3-1$. However there no linear combination of $\Lambda_1$, $\Lambda_2$ which gives $0$. Therefore, we consider second degree polynomials. The obvious choice is $J_1=x^2$, $J_2=y^2$, $J_3=xy$ with cofactors $\Lambda_1=2 -4 xy^3$, $\Lambda_2=4x^3 y-2$, $\Lambda_3=2x^3 y- 2 xy^3$ respectively. There is a linear combination which gives zero but it produces the trivial integral $1$. However, if we solve $\sum_i \mu_i \Lambda_i=-( \partial_x f_1+\partial_y f_2) $ we find the solution $\mu_1=0$, $\mu_2=0$, $\mu_3=-2$. According to the Prelle-Singer algorithm $\prod_i J_i^{\mu_i}$ is an integrating factor. This gives $(xy)^{-2}$.

I wrote a paper on this subject but unfortunately it is in Greek!

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    That looks interesting, but I'm not familiar with it! I hope I find it in my future lectures.2012-04-10
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Look for an integrating factor of the form $\mu(x,y)=\mu(x\, y)$. Taking into account that \frac{\partial\mu(x\,y)}{\partial y}=x\,\mu'(x\,y)\text{ and }\frac{\partial\mu(x\,y)}{\partial x}=y\,\mu'(x\,y) the equation $ \frac{\partial}{\partial y}\Bigl(\mu\,(2\,x^3y^2-y)\Bigr)=\frac{\partial}{\partial x}\Bigl(\mu\,(2\,x^2y^3-x)\Bigr) $ leads to \frac{\mu'(x\,y)}{\mu(x\,y)}=-\frac{2}{x\,y}. Since the right hand side depends only on $x\,y$, such an integrating factor exists. Calling $t=x\,y$, we have \mu'(t)/\mu(t)=-2/t. Form this $\mu(t)=1/t^2$. Then $\mu(x,y)=1/(x\,y)^2$ is an integrating factor. Multiply the original equation by $1/(x\,y)^2$ to get $ \Bigl(2\,x-\frac{1}{x^2y}\Bigr)dx+\Bigl(2\,y-\frac{1}{x\,y^2}\Bigr)dy=0. $ I leave it to you to check that it is exact and solve it.

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    You do the same as with integrating factors that depend only on$x$or only on $y$. I have spelled it out in he answer.2012-04-08
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$(2x^3y^2-y)dx+(2x^2y^3-x)dy=0$

$2(x^3y^2dx+x^2y^3dy)-(xdy + ydx)=0$

$(x^2y^2)*2(xdx+ydy) = d(xy)$

$d(x^2+y^2)=\frac{d(xy)}{(xy)^2}$

which upon Integration gives

$x^2+y^2+\frac{1}{xy}=c$

Some Results:

$xdy+ydx=d(xy)$

$xdx+ydy=\frac{1}{2}d(x^2+y^2)$

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    Thanks! I deleted my comment before I saw you wrote one, because I understood your steps. Thanks any way!2012-04-09