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 $$tan^{-1}\left(y/x\right)\ne -tan^{-1}\left(x/y\right)$$

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ You missed $+C$ for indefinite integrals $\endgroup$ – Math Lover Sep 26 '17 at 19:12
  • 1
    $\begingroup$ 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

plugging $$y=Cx$$ in your equation we get $$\frac{1}{x}-\frac{Cx}{x^2+(Cx)^2}+\frac{Cx}{x^2+(Cx)^2}-\frac{1}{x}=0$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.