Question

Trying to solve an ODE, prompt posted below shortly.

Answer 1

\[\text{Show that the differential equation is homogeneous and then}\]\[\text{solve the homogeneous equation subject \to the given condition.}\]
\[(x+ye^{y/x})dx-xe^{y/x}dy=0, \ \ \ y(1)=0.\]

Answer 2

I'm not sure how I'm supposed to "show it's homogeneous", g(x) is identically zero, so what else do I need to do?

Answer 3

@Concentrationalizing

Answer 4

OH

Answer 5

See, this was confusing me, I thought it was talking about being a homogeneous DE where g(x) equals zero, this is talking about the "homogeneous to the order alpha" thing, isn't it.

Answer 6

I still am not sure how to solve these, but I'll set stuff up in a minute.

Answer 7

Yeah, so you can meet the conditions to use a substitution

Answer 8

Alright, just "plugging something in" seems really tempting, but I'm not sure how I should approach that in a more systematic manner. \[e^{y/x}\]looks great for substitution, but just putting u in its place is at odds of my idea of how this theory works.

Answer 9

@Concentrationalizing , to my knowledge, it should be something like

Answer 10

\[x^\alpha\]or\[t\]that you can factor out of it, like in this:

Answer 11

Something like this from my book: http://i.imgur.com/COivUiB.png

Answer 12

And I can't factor out e^(y/x) from both terms, only part of the first term and the whole second term, it's that standalone x that is giving me trouble.

Answer 13

There is a more systematic manner. If we can show it's homogenous. we can make the substitution y = vx, giving us also dy = vdx + xdv

Answer 14

Alright, I think I know what you're talking about vaguely although I didn't get it the first time I read it, going to reread it now: http://i.imgur.com/ET5YIfg.png

Answer 15

Its also possible to do this with x = vy and dx = vdy + ydv, which one you choose is based on which one you think is simpler to do.

Answer 16

So I need to make a substitution to *put* it into the form of a homogeneous equation, or because it is already in the form of a homogeneous equation (I don't see that), I can perform this substitution; I'm thinking the first thing I said is the correct line of causality.

Answer 17

Because it already is a homogenous equation you can do the substitution.

Answer 18

But I don't see how it is, I mean, I can't factor out a term from both equations cleanly, can I? Or, what would you factor out/propose that it's a homogeneous equation of order what?

Answer 19

e.g. how do I show that it is homogeneous?

Answer 20

Oh, wait a minute, I misread the book, I think I get how I can show how it is homogeneous, one sec.

Answer 21

\[M(x,y)dx=(x+ye^{y/x})dx; \ \ \ \]...no, nevermind, I did misread the book, but I still don't see how I can show it to be homogeneous.

Answer 22

Take \(M(tx,ty)\) and show that you can write it as \(t^\alpha M(x,y)\).
\[M(tx,ty)=tx+tye^{(ty)/(tx)}==t(x+ye^{y/x})=tM(x,y)\]

Answer 23

Unless I factor out that e^{y/x} to give the x-term a coefficient to the negative power.

So what, is t one here? Is it just one?

Answer 24

Your book doesn't explicitly say what \(t\) is supposed to denote, so I would assume it's a scalar or possibly some function that doesn't depend on \(y\) or \(x\). In any case, you're more concerned with showing that there is an \(\alpha\) such that
\[M(tx,ty)=t^\alpha M(x,y)\]
and that happens to be true with \(\alpha=0\).

Answer 25

So t can be anything at all (assuming nothing complex or crazy like infinities) in this situation since t^0 = 1?

Answer 26

And thus, it's a homogeneous function of order zero(?)

Answer 27

According to Wikipedia, \(t\) is a constant parameter: http://en.wikipedia.org/wiki/Homogeneous_differential_equation#Homogeneous_functions

And yes, homogeneous of degree 0.

Answer 28

\[(xt + yte^{yt/xt})dx -(xte^{yt/xt})dy\]
The t's in the exponents of e cancel out, which is why we can factor without the exponentials causing issues.
\[t[(x+ye^{y/x})dx -xe^{y/x}dy]\]
So yeah, this will equal \(tM(x,y)\). So I'd say this would be degree 1, but yeah.

Answer 29

Oops I meant 1, not 0 -_-

Answer 30

Oh, so you take a function\[f(x,y)\]And you change the input/multiply the input like such by t:\[f(xt,yt);\]You aren't necessarily just realizing that something is there by looking at it, you look at it and *introduce* something into the function itself to realize whether that property exists or not?

All instances of x and y are just multiplied by t to some given power? And if so, why not t^2, or t^3? Why t?

Answer 31

Or does the degree not really matter? You just have to prove that both simultaneously possess the same degree of alpha?

Answer 32

Pretty much. You usually can recognize, though by counting the exponents of all the variables. Like if I had
\(f(x,y) = x^{3}y - x^{2}y^{2}\)
I can see that each term has a total exponential count of 4, so this would be homogenous degree 4. But if I had:
\(f(x,y) = x^{2} + xy^{2}\)
This would not be homogenous, the first term has total exponents of 2 while the 2nd had total exponents of 3. Plugging in the t is a nice way to visualize this, though.

Answer 33

Oh, okay, give me one second.

Answer 34

So, just making sure this is explicitly answered, I understand that you can show functions to be homogeneous to a given degree, nonhomogeneous, or homogeneous in multiple degrees, like this one? Like, when I look at the question prompt, I can see it being homogeneous to multiple orders of alpha.

Answer 35

(And the importance lies only in that you can demonstrate it to be homogeneous, not really the order in and of itself)

Answer 36

The order doesnt matter as far as I know.

Answer 37

\[f(t^3x,t^3y)=(t^3x+t^3ye^{t^3y/t^3x})=t^3(x+ye^{y/x})\]

Alright, cool, thank you. So that's how I demonstrate that it's homogeneous, now the substitution itself.

Answer 38

Either \[y=ux,\]or\[x=vy.\]
Going to go with the first one, I guess. One moment.

Answer 39

\[y=ux, \ \frac{dy}{dx}=u \ dx\]

Answer 40

\[(x+uxe^{ux/x})dx-xe^{ux/x}dy=0, y(1)=0.\]

Answer 41

(I need to mess with one of those differentials, haven't yet thought about that, one minute.

Answer 42

\[(x+uxe^{u}) \ dx-xe^u \ dy=0.\]

Answer 43

You have to consider the u as a variable. It needs to be dy = xdu + udx

Answer 44

Alright, cool. Thank you.

Answer 45

So this whole thing becomes, \[(x+uxe^u)dx+(xe^{u})[xdu+udx]=0\]

Answer 46

And theeeennnnn......lol, one sec.

Answer 47

It's been reduced to a separable first-order DE(?)

Answer 48

You still have to find a way to separate it, lol.

Answer 49

Yeah, now it's just algebra.

Answer 50

(I know I had to keep going, heh, just had to check my book, right now I'm at the stage where I'm half-unsure how to proceed half of the time.)

Answer 51

Gotcha. Well, as long as you get the idea of what we're trying to do and why.

Answer 52

Alright, cleared my history cache, OS showing up again, hooray.

(I'm still not sure why that works.)