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$\begingroup$

$$xdx+ydy=\frac{xdy-ydx}{x^2+y^2}$$

$$(x^2+y^2)(xdx+ydy)={xdy-ydx}$$

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

$\frac{\partial M}{\partial y}=2xy+1\neq 2xy-1 =\frac{\partial N}{\partial x}$

  1. $\frac{M_{y}-N_{x}}{M}=\frac{2}{x^3+y^2x+y}\neq h(y)$

  2. $\frac{N_{x}-M_{y}}{N}=\frac{-2}{x^2y+y^3-x}\neq g(x)$

  3. $\frac{M_{y}-N_{x}}{y\cdot N-x\cdot M}= \frac{2}{y^4-2xy-x^4}\neq k(xy)$

How should I continue from here?

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    The rhs is the derivative of arctan(y/x) – 2017-01-31
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    @Shobhit which RHS? – 2017-01-31
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    There is certain discrepancy in your very second line. – 2017-01-31
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    The first line, your question – 2017-01-31
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    @Resorcinol sorry, fixed – 2017-01-31
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    Simply use: http://math.stackexchange.com/questions/2108487/simplify-int-udv-int-vdu/2108492#2108492 – 2017-01-31
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    @MyGlasses So I get $xy=arctan(\frac{y}{x})+C$? – 2017-01-31
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    @MyGlasses the answer that given is $\frac{x^2+y^2}{2}+arctan(\frac{x} {y})=C$ it seems to be $U$ of the exact ODE – 2017-01-31

3 Answers 3

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$$xdx+ydy=\frac{xdy-ydx}{x^2+y^2}$$ $$\frac12d(x^2)+\frac12d(y^2)=d\arctan\frac{y}{x}$$ $$\frac12(x^2+y^2)=\arctan\frac{y}{x}+C$$

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    Someone flagged this answer. But I found it correct.--) – 2017-01-31
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    What I mean is that your answer is correct.. With that I up vote--:) – 2017-01-31
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You have better leave the original equation and integrate each side. Both are (locally) integrable. The best is perhaps to use polar coordinates, if you are permitted to change variables: $x=r\cos \phi, y=r\sin \phi$ yields the equation: $rdr = d\phi$.

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    currently looking for an answer without change of variables – 2017-01-31
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Hint. Using polar coordinates $x=r\cos \theta,\;y=r\sin \theta$:

$$xdx+ydy=\ldots=-rdr,\quad xdy-ydx=\ldots=r^2d\theta$$ so, the new equation is $-rdr=d\theta.$