Which solid figure has one polygon base?
2 answers:
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I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take 
, so that 
and we're left with the ODE linear in 
:
Now suppose
has a power series expansion
Then the ODE can be written as
![\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge2%7D%5Cbigg%5Bn%28n-1%29a_n-%28n-1%29a_%7Bn-1%7D%5Cbigg%5Dx%5E%7Bn-2%7D%3D0)
All the coefficients of the series vanish, and setting
in the power series forms for and 
tell us that 
and 
, so we get the recurrence
We can solve explicitly for
quite easily:
and so on. Continuing in this way we end up with
so that the solution to the ODE is
We also require the solution to satisfy, which we can do easily by adding and subtracting a constant as needed:
Now suppose
Then the ODE can be written as
All the coefficients of the series vanish, and setting
We can solve explicitly for
and so on. Continuing in this way we end up with
so that the solution to the ODE is
We also require the solution to satisfy
For this case we have:
By properties of the radicals
So:
.
Now, for power properties we have:
Thus,
So:
in its radical form
Answer:
in its simplest form.
in its radical form