How to prove that $\frac{1}{xP(x,y)+yQ(x,y)}$ is an integrating factor of a homogeneous linear differential equation $P(x,y)$d$x+Q(x,y)$d$y=0$?
I have seen a proof:
Suppose that $Q \neq 0$, multiply both sides of the equation by $\frac{1}{Q}$, then:
$$\frac{\text{d}y}{\text{d}x}=-\frac{P}{Q}=-g(\frac{y}{x})$$
$$g(\frac{y}{x})\text{d}x+\text{d}y=0$$
(Here is the first place that I cannot understand, why does it equal $-g(\frac{y}{x})$.)
Let $y=ux$, then $\text{d}y=u\text{d}x+x\text{d}u$, then
$$(g(u)+u)\text{d}x+x\text{d}u=0$$
Multiply both sides of the equation by $\frac{1}{x}$, we get
$$\frac{g(u)+u}{x}\text{d}x+\text{d}u=0$$
(Also I don't understand the reason why it multiplies $\frac{1}{x}$, how can we know that it simplifies the calculation.)
The equation has an integrating factor with respect only to $u$ since
$$\dfrac{\dfrac{\partial\dfrac{g(u)+u}{x}}{\partial u}}{-\dfrac{g(u)+u}{x}}=-\dfrac{g^{\prime}(u)+1}{g(u)+u}$$
where denominator the is nonzero, let it be $\lambda(u)=\frac{1}{g(u)+u}$.
Therefore the initial equation has an integrating factor in form:
$$\mu(x,y)=\frac{1}{Q}\frac{1}{x}\lambda(u)=\frac{1}{xP(x,y)+yQ(x,y)}$$
with $xP(x,y)+yQ(x,y)\neq 0$.
And is there any easier way to understand the integrating factor?