Question

Why is the following a homogeneous differential equation of degree two?:

x sin y/x (y dx + x dy) + cos y/x (x dy - y dx) = 0

Answer 1

if you can write it in the form such that y/x is a simplified version; then it can be veed out i think

Answer 2

maybe because you have two variables in each monomial? dunno. might be...

Answer 3

degree 2 would imply a ^2 someplace

Answer 4

degree two means derivative of power 2 in the case of differential equations. I don't see a second order derivative around as for the homogeneity, 2 secs.

Answer 5

its homo becasue its got a =0 on the end

Answer 6

amistre64, are you sure about the =0 thing you just said?

Answer 7

they use 2 defs for homogenous in diffy qs; this one conforms to the =0 def

Answer 8

what's the other def?

Answer 9

f(tx,ty) = t f(x,y)

Answer 10

o that's the order 0 homo eq, right?

Answer 11

if you attach a random variable such as "t" to all your xs and ys; and can factor it out cleanly; its homogenous

Answer 12

i got no idea what the name is lol

Answer 13

it's alright, wikipedia said it, i was asking for confirmation

Answer 14

:)

Answer 15

if wiki says it; it MUST be true lol

Answer 16

lol

Answer 17

it's not as unreliable as people think, in my opinion

Answer 18

there is a (small) team of people constantly watching excluding the rest of the world

Answer 19

they say a chain is only as strong as its weakest link ...

but yeah, ive never found anything glaringly false on it yet

Answer 20

they also say a bodybuilder is as strong as his weakest muscle lol

Answer 21

but ya, why is it degree two?

Answer 22

can you algebra it into a ^2 on it somplace?

Answer 23

well i can make x^2 but that's not a derivative.

Answer 24

like x^2 vs d^2 x/dt is not the same thing

Answer 25

(random example)

Answer 26

http://www.wolframalpha.com/input/?i=x+sin+y%2Fx+%28y+dx+%2B+x+dy%29+%2B+cos+y%2Fx+%28x+dy+-+y+dx%29+%3D+0

th ewolf gets confused lol

Answer 27

lol, i'll see if i can make it understand

Answer 28

degree is not order;

d^2/dx^2 is an order, not a degree

Answer 29

ooo

Answer 30

but wait then i'd need y^2 rather than x^2, no?

Answer 31

http://ocw.stut.edu.tw/file.php/14/Ordinary_Differential_Equations/%E8%AC%9B%E7%BE%A9/Ode1.pdf

Answer 32

its not a wiki, but its got an edu at the end ;)

Answer 33

I was able to get it into the form dx/dy=(something w/ x^2 in it)
so I don't see why it has to be y^2

Answer 34

maybe multiply thru by d/dx

Answer 35

um, thats just taking the derivative of it tho, maybe lol

\[x sin \frac yx (y dx + x dy) + cos \frac yx (x dy - y dx) = 0\]
\[x sin \frac yx (y + x \frac{dy}{dx}) + cos \frac yx (x\frac{dy}{dx} - y) = 0\]
\[ yx sin \frac yx + xx sin \frac yx \frac{dy}{dx} + xcos \frac yx\frac{dy}{dx} - ycos \frac yx = 0\]
\[ yx sin \frac yx + \frac{dy}{dx} (x^2 sin \frac yx + xcos \frac yx) - ycos \frac yx = 0\]

Answer 36

ya d/dx is just taking the derivative ;) and still, all i see is x^2

Answer 37

o i just saw turingtest's answer, well: y is the dependent variable, x is independent. that doesn't matter?

Answer 38

It says that in the problem?

Answer 39

What? That's it's homogeneous of degree two? Yes, the solution says that.

Answer 40

no, that y is the dependent variable

Answer 41

No, it doesn't but it seems to reat it as the dependent variable especially with the y = xv substitution.

Answer 42

are you sure youve notated it correctly?

Answer 43

treat* and what do you mean "notated". do you mean copied it down correctly?

Answer 44

yes, to notate is to write down

Answer 45

I'm looking at something that says the degree is the exponent on the derivative, so we seem to be looking for\[(\frac{dy}{dx})^2\]somehow...

Answer 46

and i just double checked and ya, i copied it down correctly?

Answer 47

TuringTest: that's order i think (not degree)

Answer 48

no, thats degree

Answer 49

\[\frac{d^2y}{dx^2};order\ 2\]
\[\left(\frac{dy}{dx}\right)^2;degree\ 2\]

Answer 50

yes, I meant the second one there^

Answer 51

wolfram apparently wants money to work now so its not going to be helpful

Answer 52

no, i think it's just for new features.

Answer 53

there's always open source software like maxima

Answer 54

new features like; "take your time and solve this" feature lol

Answer 55

lol

Answer 56

\[yx sin \frac yx + \frac{dy}{dx} (x^2 sin \frac yx + xcos \frac yx) - ycos \frac yx = 0\]
\[y=vx;\ \frac{dy}{dx}=\frac{dv}{dx}x+v\]
\[vx^2 sin (v) + (\frac{dv}{dx}x+v) (x^2 sin (v) + xcos (v)) - vxcos(v) = 0\]
\[vx^2 sin (v) + \frac{dv}{dx}x^3sin(v)+\frac{dv}{dx}x^2cos(v)+vx^2sin(v)+vxcos(v) - vxcos(v) = 0\]
\[vx^2 sin (v) + \frac{dv}{dx}(x^3sin(v)+x^2cos(v))+vx^2sin(v)+vxcos(v) - vxcos(v) = 0\]
still cant get a deg2

Answer 57

\[\frac{dv}{dx}(x^3sin(v)+x^2cos(v))+2vx^2sin(v) = 0\]
\[\frac{dv}{dx}(x^3sin(v)+x^2cos(v))=-2vx^2sin(v)\]
\[\frac{dv}{dx}=-2\frac{vx^2sin(v)}{x^3sin(v)+x^2cos(v)}\]
and cant seem to factor out a x^n from the underside to make it homogenous eaither

Answer 58

The book implies it's of second degree prior to the substitution.

Answer 59

my book says that a dependant set of vectors is independant; so I have to kinda have to wonder how much faith to put into book stuff

Answer 60

so, what does that mean for my problem?

Answer 61

it means that we cant seem to adapt the problem and conform it to the information given of homogenous and dergree 2.

either the solution is : not exist

or

its simply a bad type up in the book

Answer 62

ugh, we also got another definition floating out there:

f(tx,ty) = t^n f(x,y) is a homogenous eg of degree "n"

Answer 63

http://www.cliffsnotes.com/study_guide/First-Order-Homogeneous-Equations.topicArticleId-19736,articleId-19713.html

Answer 64

so, attach a t to all your xs and ys and see if we can factor it our cleanly

Answer 65

\[tx sin ty/tx (ty dx + tx dy) + cos ty/tx (tx dy - ty dx) = 0\]
\[tx sin y/x (ty dx + tx dy) + cos y/x (tx dy - ty dx) = 0\]
\[tx sin y/x t(y dx + x dy) + cos y/x t(x dy - y dx) = 0\]
\[t^2 x sin y/x (y dx + x dy) + tcos y/x (x dy - y dx) = 0\]
\[t(t x sin y/x (y dx + x dy) + cos y/x (x dy - y dx) = 0\]

we are still stuck with a "t" inside

Answer 66

you sure you typed it up correctly?

Answer 67

there should be a:

y cos y/x

or maybe even

x cos y/x

to attach another t onto to pull it out clean

Answer 68

or if that: x sin y/x was just sin y/x but then thats just a t^1 and not a t^2 that gets factored out

Answer 69

Oh, I see. The problem seems well solved so it might be just this part with a typo. Thanks for all this and sorry for exhausting you with "proofs." Yes, I double checked what I typed.

Answer 70

good luck :)

Answer 71

Thanks again :).