Question

The solution of $\frac{xdy}{x^2+y^2}=(\frac{y}{x^2+y^2}-1)dx$ is

Answer 1

$\begin{array}{c}\frac{xdy}{x^2+y^2}=\left(\frac{y}{x^2+y^2}-1\right)dx\\ M\left(x,y\right)=\frac{y}{x^2+y^2}-1\\ \frac{\partial m}{\partial y}=\frac{\left(x^2+y^2\right)\cdot --y\cdot 2y}{{\left(x^2+y^2\right)}^2}\end{array}$

Now,

$\begin{array}{c}N\left(x,y\right)=\frac{-x}{x^2+y^2}\\ \frac{\partial N}{\partial x}=-\left(\frac{x^2+y^2-x\cdot 2x}{{\left(x^2+y^2\right)}^2}\right)\\ =\frac{x^2+y^2}{{\left(x^2+y^2\right)}^2}\\ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\\ \end{array}$

It is an exact equation,

$\begin{array}{c}\int \left(\frac{y}{x^2+y}-1\right)dx-\frac{xdy}{x^2+y^2}=0\\ \int M\left(x,y\right)dx=\int \left(\frac{y}{x^2+y}-1\right)\\ =y\cdot \frac{1}{y}{\tan }^{-1}\left(\frac{x}{y}\right)-x\\ ={\tan }^{-1}\left(\frac{x}{y}\right)-x\end{array}$

Now,

$\begin{array}{c}\int N\left(x,y\right)dx=\int \frac{-x}{x^2+y}dy\\ =-x\cdot \frac{1}{x}{\tan }^{-1}\left(y/x\right)\\ =-{\tan }^{-1}\left(y/x\right)\\ f\left(x,y\right)={\tan }^{-1}\left(x/y\right)-x-{\tan }^{-1}y/x\end{array}$

General Soiution,

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