Can you find this tough Integrating Factor? $$ u\left(du − dv\right) + v\left(du + dv\right) = 0 $$
Can you get this equation in a form by which the integrating factor can be found using the standard algorithm?
Expression $\;\displaystyle \mu = \frac1{u^2 + v^2}\;\: $ is given as the solution
 A: Integrating Factor
This equation is equivalent to
$$
(u+v)\,\mathrm{d}u-(u-v)\,\mathrm{d}v=0\tag{1}
$$
The integrating factor should be a function $g$ so that when multiplied by $(1)$ gives an exact differential. This is when
$$
\frac{\partial}{\partial v}[(u+v)g]=\frac{\partial}{\partial u}[-(u-v)g]\tag{2}
$$
$(2)$ implies
$$
(u-v,u+v)\cdot\nabla\log(g)=-2\tag{3}
$$
which, when normalized, becomes
$$
\frac{(u-v,u+v)}{\sqrt{2u^2+2v^2}}\cdot\nabla\log(g)=-\frac{\sqrt2}{\sqrt{u^2+v^2}}\tag{4}
$$
The left side of $(4)$ is the directional derivative, $45^\circ$ counterclockwise from radial, of $\log(g)$. We can take $g$ to be a radial function, then $(4)$ becomes
$$
\frac1{\sqrt2}\frac{\mathrm{d}}{\mathrm{d}r}\log(g)=-\frac{\sqrt2}r\tag{5}
$$
which is solved by integration as
$$
\begin{align}
g
&=\frac1{r^2}\\
&=\frac1{u^2+v^2}\tag{6}
\end{align}
$$

Solution
$(6)$ says that the left side of
$$
\frac{(u+v)\,\mathrm{d}u-(u-v)\,\mathrm{d}v}{u^2+v^2}=0\tag{7}
$$
is exact. Therefore, to solve it, we can simply integrate along any path. Integrating $(7)$ along a rectilinear path (one in which $u$ is held constant while $v$ varies, and vice-versa) gives
$$
\log\left(u^2+v^2\right)-2\tan^{-1}\left(\frac vu\right)=C\tag{8}
$$
which means that parametrically
$$
(u,v)=ce^\theta(\cos(\theta),\sin(\theta))\tag{9}
$$
A: $$u(du-dv)+v(du+dv)=0$$
Solving with integrating factor isn't the simplest method in the present case. But it is the required method. So, we have to do it.
$$(u+v)du+(-u+v)dv=0$$
We look for an integrating factor $f(u,v)$ so that we get the exact differential of a function $F(u,v)$ :
$$f(u,v)\left((u+v)du+(-u+v)dv)\right)=dF(u,v)$$
$$\frac{\partial F}{\partial u}=(u+v)f\quad\text{and}\quad \frac{\partial F}{\partial v}=(-u+v)f$$
$$\frac{\partial^2 F}{\partial u\partial v}=\frac{\partial (u+v)f}{\partial v}=\frac{\partial (-u+v)f}{\partial u}=f+(u+v)\frac{\partial f}{\partial v}=-f+(-u+v)\frac{\partial f}{\partial u}$$
$$(-u+v)\frac{\partial f}{\partial u}-(u+v)\frac{\partial f}{\partial v}=2f$$
This PDE can be solved thanks to the method of characteristics :
The set of characteristic equations is :
$$\frac{du}{-u+v}=\frac{dv}{-(u+v)}=\frac{df}{2f}$$
Using the wellknown identity $\frac{A}{B}=\frac{C}{D}=\frac{\alpha_1A+\alpha_2C}{\alpha_1B+\alpha_2D}$
$\frac{du}{-u+v}=\frac{dv}{-(u+v)}=\frac{udu+vdv}{u(-u+v)-v(u+v)}=\frac{udu+vdv}{-u^2-v^2}$
From this, a first characteristic curve is obtained (for the present purpose, we don't need for a second characteristic cuve, since we are not looking for the general solution of the PDE) : 
$$-\frac{1}{2}\frac{d(u^2+v^2)}{u^2+v^2}=\frac{df}{2f}$$
$$-\ln(u^2+v^2)=\ln(f)+\text{constant}$$
$$f(u,v)=\frac{c}{u^2+v^2}$$
This is the expected integrating factor (doesn't matter the constant).
Then, solving the initial ODE is straightforward :
$$\frac{\partial F}{\partial u}=(u+v)\frac{1}{u^2+v^2}\quad\text{and}\quad \frac{\partial F}{\partial v}=(-u+v)\frac{1}{u^2+v^2}$$
$$F(u,v)=\int_{v=\text{constant}} \frac{u+v}{u^2+v^2}du =\int_{u=\text{constant}} \frac{-u+v}{u^2+v^2}dv $$
$$F(u,v)= \frac{1}{2}\ln(u^2+v^2)-\tan^{-1}\frac{v}{u}+\text{constant}$$
Since $dF=0 \quad\to\quad F=$constant, the solution of the ODE expressed on implicit form is :
$$\frac{1}{2}\ln(u^2+v^2)-\tan^{-1}\frac{v}{u}=C$$
