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I have this as homework:

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

I tried to solve it by substituting $z=xy+1$, but got the answer like $y=Cxe^{xy}$, which, I guess, is wrong.

I tried to solve it couple of times, but with no success.

P.S. I'm new on math.stackexchange.com. Do I need to show the whole process?

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2 Answers

up vote 2 down vote accepted

Your answer $y = Cxe^{xy}$ is perfectly correct. It is, of course, an implicit solution rather than an explicit one. In order to get $y$ explicitly as a function of $x$ you would need the Lambert W function.

To check that your answer is correct, write it as $C = \dfrac{x}{y} e^{xy}$ and take the differential of both sides.

$$0 = \dfrac{\partial}{\partial x} \left(\dfrac{x}{y} e^{xy}\right) \ dx + \dfrac{\partial}{\partial y} \left(\dfrac{x}{y} e^{xy}\right) \ dy $$ Then see that this simplifies to $0=0$ using your differential equation.

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Thank you very much! That is it. I spent about 3 hours trying to get what is wrong, solving again and again. I didn't notice, the asked solution may be implicit. Anyway, I am glad to know I was right. Thank you very much again! –  Michael Sazonov May 21 '12 at 18:04
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This equation is a typical one to be solved using the integrating multiplier. Thereby you can check the validity of your answer as well. Break it down in two parts: $$xy^{2}dx+x^{2}ydy=0$$ $$ydx-xdy=0$$ For the first part it is (relatively) easy to see that the integrating multiplier is $\mu_{1}=\frac{1}{x^{2}y^{2}}$ leading to solution $xy=C_1$, hence the most general form of the multiplier shall be: $$\mu_{1}=\frac{1}{x^{2}y^{2}}\psi_{1}\left(xy\right)$$ where $\psi_1$ is some arbitrary function It can also be seen that the second part has integrating multiplier $\mu_{2}=\frac{1}{x^{2}}$ leading to solution $\frac{x}{y}=C_2$. Hence the most general form is $$\mu_{2}=\frac{1}{x^{2}}\psi_{2}\left(\frac{x}{y}\right)$$ Now setting $\psi_1(t)=t$ and $\psi_2(t)=t$ we arrive at a form of the integrating multiplier that will suit both: $$\mu=\frac{1}{xy}$$ Applying it to the given equation we obtain: $$\left(y+\frac{1}{x}\right)dx+\left(x-\frac{1}{y}\right)dy=0$$ Rearranging: $$d\left(xy\right)+d\left(\ln\frac{x}{y}\right)=0$$ Leading to the same solution Method is borrowed from the book "A Course of Differential Equations" by V.V. Stepanov

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