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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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1  
Try and exploit the symmetry. –  Pedro Tamaroff Apr 8 '12 at 20:29
    
Solutions that I've found $y=x$, $x=0$ –  no identity Apr 8 '12 at 20:37

3 Answers 3

up vote 2 down vote accepted

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

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

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

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

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 for pointing that out, that was a typing mistake which no longer exists, I have edited the question to include the 2 identities I have used , hope that helps. And welcome, :-) –  Tomarinator Apr 9 '12 at 9:17
    
Thanks! I deleted my comment before I saw you wrote one, because I understood your steps. Thanks any way! –  Jozef Apr 9 '12 at 9:19

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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Thank you for the answer. What do you with this kind of integrating factor? I study Ordinary differential equation 1, this type of integrating factor is not familiar to me. –  Jozef Apr 8 '12 at 21:03
    
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. –  Julián Aguirre Apr 8 '12 at 21:05
1  
Thanks a lot Julian. –  Jozef Apr 8 '12 at 21:22

$\mu(x,y)=\frac{1}{x^2 y^2}$ 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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Why don't you add a little more to your answer? It seems the OP is not too illustrated in this topic. –  Pedro Tamaroff Apr 8 '12 at 22:09
    
That looks interesting, but I'm not familiar with it! I hope I find it in my future lectures. –  Pedro Tamaroff Apr 10 '12 at 0:12

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