# Homework Help: Quick DE fix

1. Apr 22, 2010

### cahill8

1. The problem statement, all variables and given/known data
I'm looking to solve a DE for a problem I have but I haven't done so in a while and must be forgetting something. I've checked the answer in Mathematica and what I'm getting is wrong.

u''(r)+1/r u'=G/m

3. The attempt at a solution

homogeneous solution:

a^2+1/r a = 0 -> a = 0, -1/r

u1 = Ae^(-1)+B

try u2 = c r (as a constant B already exists as part of the solution, multiply by r)

u' = c u'' = 0

c/r = G/m
c = Gr/m

u2 = Gr^2/m

u = Ae^(-1)+B+Gr^2/m

Whereas the actual solution is

u = A Ln(r)+B+Gr^2/4m

Thanks for any help

2. Apr 22, 2010

### Staff: Mentor

You are tacitly assuming that r is a constant, but it's not. It's the independent variable.

Your DE is u''(r) + (1/r)u'(r) = G/m
Since there's no u(r) term, I would make this substitution.
Let w(r) = u'(r) ==> w'(r) = u''(r)
With this substitution you can rewrite the DE as
w'(r) + (1/r)w(r) = G/m
Now you can use the standard technique of finding an integrating factor, v(r) = e^{integral(1/r).
That's what I would do.

3. Apr 22, 2010

### cahill8

Thanks, that worked out good.

Could you point out in my initial work where I implied r was constant? because I didn't mean to.

4. Apr 22, 2010

### Staff: Mentor

You treated the equation u'' + (1/r)u' = G/m as if it were a constant coefficient differential equation.

From the homogeneous equation, u'' + (1/r)u' = 0, you got a characteristic equation of a^2 + (1/r)a = 0, and then factored the left side. This would have worked if r were a constant, but r is the independent variable, or so I gather by your use of u''(r).

It's the difference between, say y' + 2y = 0 (with general solution y = Ae-2x) and y' + xy = 0 (with general solution y = Ce-(1/2)x2).

5. Apr 22, 2010

### cahill8

Cheers, I understand now