How should I apply the homogeneous condition to check $\mu$ to be an integral factor?

Question

Assume that $$ M(x,y)dx+N(x,y)dy=0 $$ is a homogeneous equation, and that $xM+yN \neq0$. Verify that $$ \mu=\frac{1}{xM+yN} $$ is an integral factor.
I've done some calculation: $$ \frac{\partial(\mu M)}{\partial y}=\frac{-x\frac{\partial{M}}{\partial{y}}-N-y\frac{\partial{N}}{\partial{y}}}{(xM+yN)^{2}}M+\mu\frac{\partial{M}}{\partial{y}} \\ \frac{\partial(\mu N)}{\partial x}=\frac{-M-x\frac{\partial{M}}{\partial{x}}-y\frac{\partial{N}}{\partial{x}}}{(xM+yN)^{2}}N+\mu\frac{\partial{N}}{\partial{x}} \\ \frac{\partial(\mu M)}{\partial y}-\frac{\partial(\mu N)}{\partial x}=\frac{xN\frac{\partial{M}}{\partial{x}}-yM\frac{\partial{N}}{\partial{y}}+yN\frac{\partial{M}}{\partial{y}}-xM\frac{\partial{N}}{\partial{x}}}{(xM+yN)^{2}} $$ I want to prove that $\frac{\partial(\mu M)}{\partial y}-\frac{\partial(\mu N)}{\partial x}=0$, but got stuck. I think I cannot go furthur since I don't know how to apply the homogeneous condition here. Thank you in advance.

Answer 1

As I have commented \begin{align} \frac{\partial(\mu M)}{\partial y}-\frac{\partial(\mu N)}{\partial x} &=\frac{xN\frac{\partial{M}}{\partial{x}}-yM\frac{\partial{N}}{\partial{y}}+yN\frac{\partial{M}}{\partial{y}}-xM\frac{\partial{N}}{\partial{x}}}{(xM+yN)^{2}}\\ &= \frac{N^2}{(xM+yN)^{2}}( x\ \partial_x + y\ \partial_y)\frac{M}{N} \\ &= \frac{N^2}{(xM+yN)^{2}}(\mathbf x \cdot\nabla)\frac{M}{N} \end{align}

Euler's homogeneous function theorem states for a homogeneous function of degree $k$ we have $$ \mathbf x \cdot\nabla f(\mathbf x) = kf(\mathbf x) $$ Since both the functions $M$ and $N$ are assumed to be of the same degree (by defintion of the homogeneous equation), then their quotient is homogeneous of degree zero (check it!). It follows that $$ \mathbf x \cdot\nabla\frac{M}{N}(\mathbf x) = 0 $$

Answer 2

$$M(x,y)dx+N(x,y)dy=0$$ Note that: $$y=tx \implies dy=xdt+tdx$$ $$(M+Nt)dx+Nxdt=0$$ Multiply by integrating factor: $$\dfrac {dx}x+\dfrac {N}{M+Nt}dt=0$$ Note that since $M,N$ are homogeneous functions: $$\dfrac {dx}x+\dfrac {N(x,y)}{M(x,y)+tN(x,y)}dt=0$$ $$\dfrac {dx}x+\dfrac {N(1,t)}{M(1,t)+tN(1,t)}dt=0$$ $$\implies f(x)dx+g(t)dt=0$$ This is integrable: $$\implies F(x)+G(t)=C$$ Note that you are asked to check the given integrating factor $\mu$ not to find it.

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$N$ is homogeneous so we can do this: $$N(x,y)=N(x,tx)=xN(1,t)$$ Do the same for the function $M$. I considered the functions as of degree $1$ of homogeneity. But it's nearly the same for another degree.