1. Originally Posted by Chris L T521
Here is one way to test and see if its homogeneous:

We (in a sense) see that $\displaystyle N(x,y)=\frac{1-3y}{y}$ and $\displaystyle M(x,y)=\frac{2-3x}{x}$ after getting our equation in the form of $\displaystyle M(x,y)\,dx+N(x,y)\,dy=0$
I see where this is bit is from, so great!

It is homogeneous if we can observe the following:

$\displaystyle M(tx,ty)=t^{\alpha}M(x,y)$ AND $\displaystyle N(tx,ty)=t^{\alpha}N(x,y)$.
For this part, what are tx and ty? It's just that the DE in question is using y's and x's.

Is t a dummy variable or something?

2. Originally Posted by Showcase_22
I see where this is bit is from, so great!

For this part, what are tx and ty? It's just that the DE in question is using y's and x's.

Is t a dummy variable or something?
Sorta. The variable t is like a dummy variable, but its something that can help us show [analytically] if the functions M(x,y) and N(x,y) [from the DE] in question are homogeneous.

If we introduce this dummy variable t and show that if we have $\displaystyle t^{\alpha}M(x,y)$ after manipulating $\displaystyle M(tx,ty)$ we're almost done showing that its homogeneous. The last part is to show that we get $\displaystyle t^{\alpha}N(x,y)$ after manipulating $\displaystyle N(tx,ty)$ AND that the we have the same $\displaystyle \alpha$ exponents. If the exponents are different, OR we can't show that $\displaystyle M(tx,ty)=t^{\alpha}M(x,y)$ or $\displaystyle N(tx,ty)=t^{\alpha}N(x,y)$, then its not homogeneous.

Does this help?

--Chris

3. I think i'm understanding it, but there's just something else that's puzzling.

We have:

$\displaystyle -\frac{2-3x}{x}dx+\frac{3y-1}{y}dy=0$

and we're comparing this with:

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

Comparing coefficients we get:

$\displaystyle M(x,y)=-\frac{(2-3x)}{x}$ and $\displaystyle N(x,y)=\frac{3y-1}{y}$

Is this equal to:

$\displaystyle M(x)=-\frac{(2-3x)}{x}$ and $\displaystyle N(y)=\frac{3y-1}{y}$?

For the next part:

$\displaystyle M(tx,ty)=t^{\alpha}M(x,y)$

$\displaystyle \frac{-(2-3tx)}{tx}=t^\alpha \frac{-(2-3x)}{x}$

This gives:

$\displaystyle \frac{-(2-3tx)}{t}=t^\alpha (3x-2)$

$\displaystyle 3tx-2=t^{\alpha +1} (3x-2)$

and since the -2 on the LHS does not contain a t then there is no value of $\displaystyle \alpha$ for which this is true.

I haven't tried it with the y's yet since it has to satisfy both these conditions to be homogeneous. Since this one doesn't work the DE is non-homogeneous.

4. Originally Posted by Showcase_22
I think i'm understanding it, but there's just something else that's puzzling.

We have:

$\displaystyle -\frac{2-3x}{x}dx+\frac{3y-1}{y}dy=0$

and we're comparing this with:

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

Comparing coefficients we get:

$\displaystyle M(x,y)=-\frac{(2-3x)}{x}$ and $\displaystyle N(x,y)=\frac{3y-1}{y}$

Is this equal to:

$\displaystyle M(x)=-\frac{(2-3x)}{x}$ and $\displaystyle N(y)=\frac{3y-1}{y}$?

For the next part:

$\displaystyle M(tx,ty)=t^{\alpha}M(x,y)$

$\displaystyle \frac{-(2-3tx)}{tx}=t^\alpha \frac{-(2-3x)}{x}$

This gives:

$\displaystyle \frac{-(2-3tx)}{t}=t^\alpha (3x-2)$

$\displaystyle 3tx-2=t^{\alpha +1} (3x-2)$

and since the -2 on the LHS does not contain a t then there is no value of $\displaystyle \alpha$ for which this is true. Correct

I haven't tried it with the y's yet since it has to satisfy both these conditions to be homogeneous. Since this one doesn't work the DE is non-homogeneous. Correct. You will end up with the same situation with the y's, where there is no $\displaystyle \alpha$ value where this would be true.