"how can I derive this solution?"

Differentiate it?

Differentiate it?

Solved

Posted on 2009-12-17

I need to solve Poisson's equation in spherical coordinates (which is modelling something related to plasmas). The equation is:

1/r^2 d/dr(r^2 * dp(r)/dr) = 2p(r)/k^2

where k is a known constant. So the first thing seems to be to take the r^2 to the other side, and apply the product rule on the left, to get the second order ODE:

2r p' + r^2 p'' = 2r^2 p/k^2

or

p'' + (2/r)p' - (2/k^2)p = 0. (*)

How do I solve this? According to Wolfram, this has the general solution:

p(r) = c1*exp(-sqrt(2)*r/k)/r + c2*exp(sqrt(2)*r/k)/(2*sqrt(2)*r/k) *EDITED*

which indeed leads to the correct solution when I apply some known conditions, but how can I derive this solution?

Thanks very much

1/r^2 d/dr(r^2 * dp(r)/dr) = 2p(r)/k^2

where k is a known constant. So the first thing seems to be to take the r^2 to the other side, and apply the product rule on the left, to get the second order ODE:

2r p' + r^2 p'' = 2r^2 p/k^2

or

p'' + (2/r)p' - (2/k^2)p = 0. (*)

How do I solve this? According to Wolfram, this has the general solution:

p(r) = c1*exp(-sqrt(2)*r/k)/r + c2*exp(sqrt(2)*r/k)/(2*sqr

which indeed leads to the correct solution when I apply some known conditions, but how can I derive this solution?

Thanks very much

5 Comments

" p(r) = c1*exp(-sqrt(2)*r/k)/r + c2*exp(sqrt(2)*r/k)/(2*sqr

so if you differentiate it and get

" p'' + (2/r)p' - (2/k^2)p = 0. "

you thought correctly

I do too but it is often the easiest way. Integration is the inverse of differentiation and inverse operations are always harder. Consider long division, the inverse of multiplication. It is just an organized trial and error procedure. The same with integration. Poisson's equation can just be looked at as the result of an organized trial and error.

In general physics SHM equation is usually just plopped down (and checked by differentiation).

The whole practice of integration is the result of trail and errors. 1st order, 2nd order, homogeneous, non-homogeneous, separable, etc)

You may be able to get somebody to help you further (but not me)'

Note which might lead to something

If the variables cannot be separated, the method of integral transforms might help.

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