# Homogeneous and particular solution

1. May 19, 2005

### gomez

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

2. May 19, 2005

### dextercioby

Who's "V"...?

Daniel.

3. May 19, 2005

### arildno

I think it is a "she"..

4. May 19, 2005

### gomez

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

5. May 19, 2005

### dextercioby

A what...?A "she"...?

Daniel.

6. May 19, 2005

### dextercioby

Make the substitution

$$\frac{dv_{z}}{dr}=u(r)$$

Daniel.

7. May 19, 2005

### HallsofIvy

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.

Last edited by a moderator: May 19, 2005
8. May 19, 2005

### arildno

Perhaps I was wrong then..