Show that x dx+y dy+x dy-y dx=0 has an integrating factor of…

Show that $x\,dx+y\,dy+x\,dy-y\,dx=0$ has an integrating factor of the form $\mu=\mu(x^2+y^2)$, and find its general integral.

Step 1: Set up and check exactness

With $M=x-y$, $N=x+y$: $M_y=-1$, $N_x=1$, so $M_y\ne N_x$ and the equation is not exact.

Step 2: Impose the condition for $\mu(r)$, $r=x^2+y^2$

For $\mu=\mu(r)$, exactness of $\mu M\,dx+\mu N\,dy=0$ requires $\mu(M_y-N_x)=(N\cdot 2x-M\cdot2y)\mu'(r)$. Here

\[N\cdot2x-M\cdot2y=2x(x+y)-2y(x-y)=2(x^2+y^2)=2r,\]

so $\mu(-2)=2r\mu'(r)$, i.e. $\dfrac{\mu'}{\mu}=-\dfrac1r$ — a genuine function of $r$ alone, confirming such a $\mu$ exists.

Step 3: Solve for $\mu$

$\ln|\mu|=-\ln r$, so $\mu(r)=\dfrac1r=\dfrac{1}{x^2+y^2}$.

Step 4: Multiply and verify exactness

$M_1=\dfrac{x-y}{x^2+y^2}$, $N_1=\dfrac{x+y}{x^2+y^2}$. Direct differentiation gives $\partial_yM_1=\partial_xN_1=\dfrac{y^2-x^2-2xy}{(x^2+y^2)^2}$ (verified with a CAS), so the equation is now exact.

Step 5: Split into recognizable exact differentials

Write the equation as $\dfrac{x\,dx+y\,dy}{x^2+y^2}+\dfrac{x\,dy-y\,dx}{x^2+y^2}=0$. The first term is $d\!\left[\tfrac12\ln(x^2+y^2)\right]$ and the second is $d\!\left[\arctan\dfrac{y}{x}\right]$, so

\[d\!\left[\tfrac12\ln(x^2+y^2)+\arctan\tfrac{y}{x}\right]=0.\]