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I was solving this exact differential equation $$\left(\frac{1}{x}-\frac{y}{x^2+y^2}\right) dx+\left(\frac{x}{x^2+y^2}-\frac{1}{y}\right) dy=0$$ I was confused about the following integration : $$\int \frac{-y}{x^2+y^2}dx=tan^{-1}\left(y/x\right) $$ or does it have to be $$\int \frac{-y}{x^2+y^2}dx=- tan^{-1}\left(x/y\right) $$ the same for the following integration , it can take 2 values, which one is right? $$\int \frac{x}{x^2+y^2}dy=tan^{-1}\left(y/x\right) $$ $$\int \frac{x}{x^2+y^2}dy=-tan^{-1}\left(x/y\right)$$ But I think that $$tan^{-1}\left(y/x\right)\ne -tan^{-1}\left(x/y\right)$$

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    $\begingroup$ Note that :$$tan^{-1}(y/x)+tan^{-1}(x/y)=\frac{\pi}{2}$$ $\endgroup$ – Khosrotash Sep 26 '17 at 19:10
  • $\begingroup$ $$y(x)=Cx$$ is one solution $\endgroup$ – Dr. Sonnhard Graubner Sep 26 '17 at 19:10
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    $\begingroup$ You missed $+C$ for indefinite integrals $\endgroup$ – Math Lover Sep 26 '17 at 19:12
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    $\begingroup$ i see the $$C$$ control the solution and plug them in the equation $\endgroup$ – Dr. Sonnhard Graubner Sep 26 '17 at 19:15
  • $\begingroup$ @Khosrotash $\frac{\pi}{2} \text{sgn}(x/y)$ $\endgroup$ – Gribouillis Sep 26 '17 at 19:17
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plugging $$y=Cx$$ in your equation we get $$\frac{1}{x}-\frac{Cx}{x^2+(Cx)^2}+\frac{Cx}{x^2+(Cx)^2}-\frac{1}{x}=0$$

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  • $\begingroup$ But I see that we can get this solution by guessing , can we get it from the integration ? ( I mean are there any steps to get this solution? or just we guessed ! ) $\endgroup$ – MCS Sep 26 '17 at 19:30

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