Homogeneous and particular solution

[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

 

[dextercioby]

Who's "V"...?

Daniel.

 

[arildno]

Who's "V"...?

Daniel.

I think it is a "she"..

 

[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

 

[dextercioby]

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

Daniel.

 

[dextercioby]

Make the substitution

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

Daniel.

 

[HallsofIvy]

I THINK (it surely isn't clear) that the OP mean that V_z 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 r²V_z"+ rV_z'= kr². If we let x= ln r, (so that r= e^x) we can rewrite that as a linear equation with constant coefficients. Specifically, (I'm writing "y" in place of "V_z" 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= ke^2x. Integrating once, dy/dt= (k/2)e^2x+ C₁. Integrating a second time, y= (k/4)e^2x+ C₁x+ C₂. e^x= r so e^2x= r². In terms of r, y(r)= V_z(r)= (k/4)r²+ C₁ln r+ C₂.

 

[arildno]

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

Daniel.

Perhaps I was wrong then..