Show that ODE is homogeneous, but I don't think it is

[overpen57mm]

Homework Statement

Show the following differential equation is homogeneous and solve it by using an appropriate substitution

dy/dx = y/(x(1-ln(x)+ln(y))

Relevant Equations

M(x,y)dx + N(x,y)dy = 0,
af(x,y) = f(ax, ay)

Ignoring the second part of the question for now, since I think it will be more clear once I understand how this equation is homogeneous.

According to my textbook and online resources a first-order ODE is homogeneous when it can be written like so:

$$M(x,y) dx + N(x,y) dy = 0$$

and $M(x,y)$ and $N(x,y)$ are both homogeneous to the same degree.

I can write this equation in that form

$$(x(1-\ln(xy)) dy - y dx = 0$$

$$M(x,y) = x(1-\ln(xy)), N(x,y) = -y$$

certainly $N(x,y)$ is homogeneous with degree 1, $aN(x,y) = -ay = N(ax, ay)$

But I can't get $M(x,y)$ to also be homogeneous with degree 1, or even homogeneous at all

$$aM(x,y) = ax(1-ln(xy))$$

$$M(ax, ay) = ax(1-ln(a^2xy)) = ax - 2axln(a)ln(xy)$$

As you can see, it's simply a different equation altogether, the $2ln(a)$ throws the whole thing off and there's no way it's homogeneous with degree 1.

So how do I get this equation to be homogeneous, and what kind of substitution am I looking for to solve this?

 

[Mark44]

I can write this equation in that form $(x(1-\ln(xy)) dy - y dx = 0$

There's a mistake above. $1 - \ln(x) + \ln(y) = 1 - \ln(x/y)$

Also, the term "homogenous differential equation" has two different meanings, due to historical circumstances.
From the textbook I have, "A First Course in Differential Equations," by Frank G. Hagin, if the DE can be written in the form $\frac{dy}{dx} = f\left(\frac y x\right)$, it is said to be homogeneous.

Rearranging the terms of your equation, I get $\frac{dy}{dx} = \frac y x \frac 1 {1 + \ln(y/x)}$. The right side certainly is a function of y/x.

 

[MidgetDwarf]

Following Mark's post. We can also see if the given DE. Mdx+Ndy= 0 is homogeneous by seeing if M and N have the same degree of homogeneity. But the converse is not necessarily true.