Solving a Homogeneous Equation: Understanding the Concept and Steps

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SUMMARY

The discussion centers on solving the homogeneous differential equation represented as xdx + sin(y/x)(ydx - xdy) = 0. Participants clarify that the sine function does not affect the homogeneity of the equation when substituting v = y/x. The correct substitution leads to a separable equation, allowing for integration. However, confusion arises regarding the presence of derivatives and the proper handling of differentials in the context of ordinary differential equations (ODEs).

PREREQUISITES
  • Understanding of homogeneous differential equations
  • Familiarity with substitution methods in ODEs
  • Knowledge of separable equations and integration techniques
  • Basic concepts of differentials and derivatives in calculus
NEXT STEPS
  • Study the method of substitution in solving ODEs, focusing on v = y/x
  • Learn about separable differential equations and their integration
  • Explore the implications of using differentials in ODEs
  • Review the properties of homogeneous functions in differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as anyone seeking to deepen their understanding of ODE techniques and concepts.

Mastur
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Homework Statement


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

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.

\frac{dy}{y} + \frac{dv}{v} + \frac{sinv}{v^2}dv = 0

ln{x} + ln{v} + \int\frac{sinv}{v^2}dv = 0

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


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

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.

\frac{dy}{y} + \frac{dv}{v} + \frac{sinv}{v^2}dv = 0

ln{x} + ln{v} + \int\frac{sinv}{v^2}dv = 0

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


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

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
\frac{dy}{dx}= \fac{x+ ysin(y/x)}{x}= 1+ \frac{y}{x}sin(y/x)
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
\frac{dy}{dx}= \fac{x+ ysin(y/x)}{x}= 1+ \frac{y}{x}sin(y/x)
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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