Solve for n: Finding Value of n for Equation V

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utkarshakash
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Homework Statement


Find the value of n so that the equation V=r^n(3 \cos ^3 \theta -1) satisfies the relation
$$\dfrac{\partial}{\partial r} \left( r^2 \dfrac{\partial V}{\partial r} \right) + \dfrac{1}{\sin \theta}\dfrac{\partial}{\partial \theta} \left( \sin \theta \dfrac{\partial V}{\partial \theta} \right)=0$$


Homework Equations



The Attempt at a Solution


The final equation after differentiation comes out to be
$$n(n+1)(3 \cos ^3 \theta -1) = 18 \cos \theta (2\cos ^2 \theta -1) $$

What should I substitute for θ ?
 
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Why should you need to substitute something for theta?
Recheck your algebra... you final relation is saying there is no value of n that will make V satisfy the PDE.
 
Simon Bridge said:
Why should you need to substitute something for theta?
Recheck your algebra... you final relation is saying there is no value of n that will make V satisfy the PDE.


I've already checked it thrice but couldn't find any error! Can you please show me where I'm going wrong?
 
utkarshakash said:
I've already checked it thrice but couldn't find any error! Can you please show me where I'm going wrong?

How can anybody tell where you went wrong? You have not shown your work in detail.

I (or Maple) get something completely different from your result.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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