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Homogeneous Equation?

  • Thread starter Mastur
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  • #1
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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.
 

Answers and Replies

  • #2
HallsofIvy
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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.
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.
 
  • #3
256
2

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?
 
  • #4
HallsofIvy
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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.
 
  • #5
256
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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.
 
  • #6
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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.
 
  • #7
HallsofIvy
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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?
 
  • #8
256
2
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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