Would someone kindly explain to me what the logic is behind one of the steps here:
http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx
In Example 1 -
Following on from this sentence on the page
"Now, at this point we just need to start plugging in some value of $n$ and see what happens"
I get the plugging in $n's$ part, I just don't understand the steps behind writing all of the even $a's$ in terms of $a_0$ and all of the odd $a's$ in terms of $a_1$
For example they have $a_4$ = $-\dfrac{a_2}{(4)(3)}$ which they then re-write as
$a_4$ = $\dfrac{a_0}{(4)(3)(2)(1)}$
Can anyone explain how this step works?
Let me know if you need clarification, thanks.
For $n=0$, we discover $a_2 = \frac{-a_0}{(2)(1)}$. We expect this from the recurence $a_{n+2} = \frac{-a_n}{(n+2)(n+1)}$. Setting $n=2$, we get $a_{2+2} = \frac{-a_2}{(2+2)(2+1)}$. But we already know $a_2$ and plugging this in for $a_2$ we get $$a_{4} = \frac{-\frac{-a_0}{(2)(1)}}{(4)(3)} = \frac{a_0}{(4)(3)(2)(1)}$$