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.**

Is that about right?