1. ## homogeneous equations...

ok i have the equation:

dy/dx = (x-y)/x

i can see how this equation is exact and i can even see how it is linear, what i don't see is how it is homogeneous. doing the problems when they were separated by type, was much easier then when they are mixed together. if someone could explain how this problem is homogeneous i would appreciate it.

also the book mentioned something about:

g(tx,ty) = t^(alpha)*g(x,y) <---what does this mean? thanks in advance.

2. Originally Posted by slapmaxwell1
ok i have the equation:

dy/dx = (x-y)/x

i can see how this equation is exact and i can even see how it is linear, what i don't see is how it is homogeneous. doing the problems when they were separated by type, was much easier then when they are mixed together. if someone could explain how this problem is homogeneous i would appreciate it.

also the book mentioned something about:

g(tx,ty) = t^(alpha)*g(x,y) <---what does this mean? thanks in advance.
Homogeneous in this sense isn't about being set equal to 0.

This is a homogeneous function.

Example:

$f(x,y)=x^3+y^3$

$f(tx,ty)=(tx)^3+(ty)^3=t^3(x^3+y^3)=t^3f(x,y)$

Now applying this to your situation.

$xdy-(x-y)dx=0$

$M(x,y)=x \ \ \ \ N(x,y)=x-y$

$M(tx,ty)=tx=tM(x,y)$

$N(tx,ty)=tx-ty=t(x-y)=tN(x,y)$

3. ok wait a min...did you mean x dy = (1-y) dx ? you wrote x dy - (1-y)dx=0

4. Originally Posted by slapmaxwell1
ok wait a min...did you mean x dy = (1-y) dx ? you wrote x dy - (1-y)dx=0
$xdy=(x-y)dx\Rightarrow xdy-(x-y)dx=0$

5. It means that the function on the right can be written as $\displaystyle f\Big( \frac{x}{y} \Big)$ or $\displaystyle f \Big(\frac{y}{x} \Big)$

6. sorry about that i wrote it down wrong.....im good.