Find the general integral of the total differential equation…

Find the general integral of the total differential equation $\dfrac{x}{x^2+y^2}dy=\left(\dfrac{y}{x^2+y^2}-1\right)dx$.

Explanation of Method

Rewrite as $M\,dx+N\,dy=0$ with $M=1-\dfrac{y}{x^2+y^2}$, $N=\dfrac{x}{x^2+y^2}$, check exactness, and recognize the potential function as an arctangent.

Step 1: Check exactness

$M_y=N_x=\dfrac{y^2-x^2}{(x^2+y^2)^2}$, so the equation is exact.

Step 2: Find the potential function

Integrating $F_y=N=\dfrac{x}{x^2+y^2}$ with respect to $y$: $F=\arctan\dfrac{y}{x}+h(x)$. Matching $F_x=-\dfrac{y}{x^2+y^2}+h'(x)=M=1-\dfrac{y}{x^2+y^2}$ gives $h'(x)=1$, so $h(x)=x$.

Result

Final answer:

\[\boxed{\,\arctan\frac{y}{x}+x=C,\qquad C\in\mathbb{R}}\]