Question

hi everyone! Could someone please look at this diffeq I have posted and tell me how they got from the 2nd to the last step to the very last step? Very confused on where the "1" went! Thanks :o)

Answer 1

Why isn't the answer dy/dx = G(X/Y , 1) ?

Answer 2

@phi

Answer 3

It looks like they defined G(x/y) to be -M(x/y,1)/N(x/y,1)

Answer 4

so they just simply said "Let stuff = other stuff" ? That's it? No mystery?

Answer 5

I think so. Can you post a page or 2 of this section?

Answer 6

it's from Chegg...let me see what I can post...brb

Answer 7

@phi

Answer 8

They start with M (and N) that are functions of both x and y
then they replace x with x/y and y with 1 (i.e. divided by y)
when we do that, M(x/y, 1) can be thought of as a function of z= (x/y)
(the 1 is a constant).
similarly for N(x/y, 1)

the idea is that G(z)= M(x/y,1)/ N(x/y,1) is only a function of a "new" variable z (= x/y)

Answer 9

yikes...that's kinda hard to understand, the "z" stuff popping into existence for example

Answer 10

clear as mud?
Let me think about another way to say it

Answer 11

well I kinda get it, I just have to stare at it for awhile

Answer 12

question...

Answer 13

if we had a 3 dimension vector space for example like [M(x,y,z)] / [N(x,y,z)]...but we were given a problem like this one above, and it said this instead...
[M(x/q,5,z^2)] / [N(x/q,5,z^2)]...could we say that the w=x/q and t=z^2 so therefore [M(x/q,5,z^2)] / [N(x/q,5,z^2)] = G(w,5,t) = G(x/q,5,z^2) ???

Does that make sense? I mean in the original problem, the constant is all the way to the right, I have no idea the proper way to handle a constant in the middle, just my best guess...is that correct?

Answer 14

I don't remember doing those kinds of problems... If you ever run into them, please post how they are handled!

Answer 15

lol...okay...I was just making something up, but taking the time to write out my nonsense has given me enough time to better understand what you wrote, so all in all...YAY! :o)

Answer 16

one last thing...

Answer 17

M(cake/pizza,3) / N(cake/pizza,3)...z=(cake/pizza)
so...G(z)=G(cake/pizza) ?

Answer 18

yes, though with homogenous functions, we won't ever get that situation... It would be a 1 (not 3) because we divide by the second variable (y ), and that will give us 1.

Answer 19

got it! thanks phi! :o)

Answer 20

btw, another way to show a function is homogenous is if it is true that
\[ f(tx, ty) = t^n f(x,y) \]
example:
\[ f(x,y) = x^2 +y^2 \ln\left(\frac{y}{x}\right) \]
replace x with tx and y with ty, where t>0:
\[ f(tx,ty) = (tx)^2 +(ty)^2 \ln\left(\frac{ty}{tx}\right) \\
= t^2 \left( x^2+y^2 \ln\left(\frac{y}{x}\right)\right) \\
= t^2 f(x,y)
\]

Answer 21

oh yes...i'm very familiar with that "t" stuff!
last question...

Answer 22

nevermind...I answered my own question...time to close this one! thanks :o)