# Simple differential equation

1. May 14, 2007

### John O' Meara

Two concentric spherical conductors radii a and b are at potentials V1 and 0 respectively. The potential V at a distance x from thier common centre is given by $$\frac{d[x^2\frac{dV}{dx}]}{dx}=0 \\$$. Find V in terms of x, a, b and V1 ( Note V=V1 when x=a and V=0 when x=b).
I just find it difficult to get started on this, I seem not able to separate out the variables dV and x. Thanks for the help.

2. May 14, 2007

### daveb

Well, if (as you've written it) the derivative is 0, then x^2V' would be constant. That is separable.

3. May 14, 2007

### John O' Meara

Do I write down x^2V' =0, then $$x^2 \int dV =0$$

4. May 14, 2007

### arunbg

No, you write
$$x^2\frac{dV}{dx} = k(constant)$$

5. May 14, 2007

### Office_Shredder

Staff Emeritus
EDIT: Bah, I screwed up somewhere in the latex. arunbg wrote the essential part anyway

6. May 14, 2007

### John O' Meara

Is that constant k not =0? I mean the $$d[x^2\frac{dV}{dx}}] = dx \times 0 = 0 \\$$ In other words where did you get the k. Anyway I have the following $$V=k\int \frac{1}{x^2}dx \\$$.

7. May 14, 2007

### cristo

Staff Emeritus
You can't separate the original equation like that. You have $$\frac{d}{dx}\left(x^2V'\right)=0$$. Integrating both sides wrt x gives $$\int \frac{d}{dx}\left(x^2V'\right)dx=k \Rightarrow x^2V'=k$$

Your next part is correct. Now integrate that.

Last edited: May 14, 2007
8. May 14, 2007

### John O' Meara

Thanks cristo for your help.

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