Solving a Homogeneous Equation: Understanding the Concept and Steps

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Homework Help Overview

The discussion revolves around a differential equation expressed as xdx + sin(y/x)(ydx - xdy) = 0. Participants are exploring whether this equation is homogeneous, particularly due to the presence of the sine function, and are attempting to clarify the implications of their substitutions.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Some participants attempt to determine the homogeneity of the equation, questioning the role of the sine function in their substitutions. Others suggest different approaches to the substitution, particularly regarding the variables used. There is also discussion about the nature of differentials and their relation to derivatives in the context of the equation.

Discussion Status

The discussion is ongoing, with participants providing hints and suggestions for corrections. There is a recognition of the need for clarity in the substitutions and the nature of the equation, but no consensus has been reached on the best approach or understanding of the problem.

Contextual Notes

Participants express confusion regarding the equation's structure, particularly the absence of explicit derivatives, and the implications this has for solving the problem. There are references to homework constraints and the need for individual problem-solving efforts.

Mastur
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Homework Statement


[itex]xdx+sin\frac{y}{x}(ydx-xdy) = 0[/itex]

The Attempt at a Solution


Well, it's quite easy. But I'm quite confused if this is homogeneous or not, because of the sine function. This is my solution, assuming that this is a homogeneous equation.

let x = vy, dx = vdy + ydv; then substituting these values in the equation gives me,

v2ydy + vy2dv + y2sinvdv = 0. The variables now are separable.

[itex]\frac{dy}{y} + \frac{dv}{v} + \frac{sinv}{v^2}dv = 0[/itex]

[itex]ln{x} + ln{v} + \int\frac{sinv}{v^2}dv = 0[/itex]

Now, I cannot integrate the last function. I still have a lot of problems to solve, and a lot of questions to ask. I'll post others later as soon as I have tried solving them.
 
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Mastur said:

Homework Statement


[itex]xdx+sin\frac{y}{x}(ydx-xdy) = 0[/itex]

The Attempt at a Solution


Well, it's quite easy. But I'm quite confused if this is homogeneous or not, because of the sine function.
It is sin of y/x and if you change both x and y to vx and vy, vy/vx= y/x so the sine is not changed. Yes, this is a homogeneous equation

This is my solution, assuming that this is a homogeneous equation.

let x = vy, dx = vdy + ydv;
What? If you let x= vy, then v= x/y and your sin(y/x) becomes sin(1/v)! You want
v= y/x so that y= vx, dy= vdx+ xdv. You want an equation in v and x, not v and y.

then substituting these values in the equation gives me,

v2ydy + vy2dv + y2sinvdv = 0. The variables now are separable.

[itex]\frac{dy}{y} + \frac{dv}{v} + \frac{sinv}{v^2}dv = 0[/itex]

[itex]ln{x} + ln{v} + \int\frac{sinv}{v^2}dv = 0[/itex]

Now, I cannot integrate the last function. I still have a lot of problems to solve, and a lot of questions to ask. I'll post others later as soon as I have tried solving them.
 
Mastur said:

Homework Statement


[itex]xdx+sin\frac{y}{x}(ydx-xdy) = 0[/itex]

I think it's kind of disturbing that your differential equation (as stated) has no derivatives in it.

Can I state the solution I found for this equation?
 
Amok said:
I think it's kind of disturbing that your differential equation (as stated) has no derivatives in it.

Can I state the solution I found for this equation?
It has differentials which is the same thing.
If you really need to have derivatives, the equation is the same as
[tex]\frac{dy}{dx}= \fac{x+ ysin(y/x)}{x}= 1+ \frac{y}{x}sin(y/x)[/tex]
which is why the substitution v= y/x (NOT v= x/y) works.

Please give Mastur a chance to do it himself. If he makes the correction I suggested, he should be able to do it easily.
 
HallsofIvy said:
It has differentials which is the same thing.
If you really need to have derivatives, the equation is the same as
[tex]\frac{dy}{dx}= \fac{x+ ysin(y/x)}{x}= 1+ \frac{y}{x}sin(y/x)[/tex]
which is why the substitution v= y/x (NOT v= x/y) works.

Please give Mastur a chance to do it himself. If he makes the correction I suggested, he should be able to do it easily.

Yes, I know, but I think it leads to confusion since usually only separate differentials when the equation is separable.
 
Oh, I see.

I'll try to re-solve the problem tomorrow. I'm quite tired right now because of the assigned activity for our group.

Thanks for the hints.
 
Amok said:
Yes, I know, but I think it leads to confusion since usually only separate differentials when the equation is separable.
Where in the world did you get that idea?
 
HallsofIvy said:
Where in the world did you get that idea?

Well, unless you are talking about more advanced math (like differential forms, which I haven't studied, so I might be wrong) I don't think differentials have any meaning by themselves and only make sense in derivatives or integrals. So to be honest, even if you have a separable ODE you shouldn't separate the differentials (i.e. they should always appear in 'fractions'). At least that's what I've always though, correct me if I'm wrong.

I'd rather not even use Leibniz notation for ODEs.
 

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