Solving a Homogeneous Equation: Understanding the Concept and Steps

In summary: I don't think it has any significant advantages over standard notation.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. I don't think it has any significant advantages over standard notation.
  • #1
Mastur
41
0

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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
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
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?
 
  • #8
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.
 

What is a homogeneous equation?

A homogeneous equation is a type of mathematical equation where all of the terms have the same degree. This means that each term has the same number of variables raised to the same power. For example, x^2 + 2xy + y^2 = 0 is a homogeneous equation because all terms have a degree of 2.

How is a homogeneous equation different from a non-homogeneous equation?

A non-homogeneous equation is one where the terms have different degrees. This means that different terms in the equation have different numbers of variables and different powers. For example, x^2 + 2xy + y = 0 is a non-homogeneous equation because the term y has a degree of 1.

What is the solution to a homogeneous equation?

The solution to a homogeneous equation is a set of values for the variables that satisfy the equation. In other words, it is the set of values that make the equation true. This solution is often represented by a single point or set of points on a graph.

How is a homogeneous equation solved?

A homogeneous equation can be solved by using various methods such as substitution, elimination, or graphing. The specific method used will depend on the type of homogeneous equation and its degree. It is also possible for a homogeneous equation to have an infinite number of solutions.

What are some real-life applications of homogeneous equations?

Homogeneous equations have many real-life applications in fields such as physics, chemistry, and engineering. For example, they can be used to model the behavior of gases, chemical reactions, and electric circuits. In physics, homogeneous equations are used to describe the motion of objects under the influence of a net force. In engineering, they are used to design structures and systems that are in equilibrium.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
575
  • Calculus and Beyond Homework Help
Replies
7
Views
706
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
474
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
Replies
12
Views
379
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
Back
Top