Homogeneous and particular solution

In summary, The conversation is about solving an ODE (Ordinary Differential Equation) with the given form: Vz''+1/r*Vz'= k, where k is a constant. The variable Vz represents velocity in the z direction and changes with respect to the independent variable r. The conversation involves making a substitution and rewriting the equation in terms of a linear equation with constant coefficients. The final solution is given by y(r)= Vz(r)= (k/4)r2+ C1ln r+ C2, where C1 and C2 are constants.
  • #1
gomez
6
0
hi,

I have this ODE and I need to obtain the general and the particular solution, this is the ODE

Vz''+1/r*Vz'= k

where k is a constant

thanks
 
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  • #3
dextercioby said:
Who's "V"...?

Daniel.
I think it is a "she"..
 
  • #4
Vz is just the velocity in the z direction, It's nothing more than my variable. Vz changed with respect to r.

Vz(r)''+1/r*Vz(r)' = k
 
  • #6
Make the substitution

[tex] \frac{dv_{z}}{dr}=u(r) [/tex]

Daniel.
 
  • #7
I THINK (it surely isn't clear) that the OP mean that Vz is the function to be solved for.

Assuming also that r is the independent variable, we can multiply the entire equation by r2 to get the "Euler-type" (or "equipotential") equation r2Vz"+ rVz'= kr2. If we let x= ln r, (so that r= ex) we can rewrite that as a linear equation with constant coefficients. Specifically, (I'm writing "y" in place of "Vz" just because it is easier) dy/dr= dy/dx dx/dr= (1/r)(dy/dx) and d2y/dr2= d/dr((1/r)dy/dx)= (-1/r2(dy/dx)+ (1/r2)d2y/dx2.
The equation becomes d2y/dx2= ke2x. Integrating once, dy/dt= (k/2)e2x+ C1. Integrating a second time, y= (k/4)e2x+ C1x+ C2. ex= r so e2x= r2. In terms of r, y(r)= Vz(r)= (k/4)r2+ C1ln r+ C2.
 
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  • #8
dextercioby said:
A what...?A "she"...?:confused:

Daniel.
Perhaps I was wrong then..
 

Related to Homogeneous and particular solution

1. What is the difference between homogeneous and particular solutions?

The main difference between homogeneous and particular solutions is that homogeneous solutions satisfy the homogeneous equation, where the right-hand side is equal to zero, while particular solutions satisfy the non-homogeneous equation, where the right-hand side is not equal to zero.

2. How do you find the homogeneous solution of a differential equation?

To find the homogeneous solution of a differential equation, you first need to set the right-hand side of the equation to zero. Then, you can solve for the unknown variables using techniques such as separation of variables or integrating factors.

3. What is the role of initial conditions in finding the particular solution?

Initial conditions are values given at a specific point in time that help determine the particular solution of a differential equation. These conditions provide the necessary information to solve for the arbitrary constants in the particular solution.

4. Can a homogeneous solution also be a particular solution?

No, a homogeneous solution and a particular solution are mutually exclusive. A homogeneous solution satisfies the homogeneous equation, while a particular solution satisfies the non-homogeneous equation.

5. Are there any real-life applications of homogeneous and particular solutions?

Homogeneous and particular solutions are widely used in various fields of science, such as physics, chemistry, and engineering. They are particularly useful in modeling and predicting natural phenomena, such as population growth, chemical reactions, and electrical circuits.

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