Question

ok

Answer 1

K.

Answer 2

By having Jalebi :3

Answer 3

O.o

Answer 4

I do believe that this looks a little bit like our good friend Airy, am I right about that?
\[\Large y''-xy=0 \]

Answer 5

Did you already come up to this point?
\[\Large \sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\sum_{n=0}^\infty a_{n}x^{n+1} \]
setting the dummy variable for n+2 on the left hand side and n-1 on the right hand side we get:
\[\Large \sum_{n=0}^\infty (n+2)(n+1)a_{n+1}x^n=\sum_{n=1}^\infty a_{n-1}x^n \]

Answer 6

Now, you can rewrite the left hand side to make it start by n=1 just by adding the first term to it, because you're "missing" it when you start it at n=1 rather than n=0
\[\Large2 a_2+\sum_{n=1}^\infty (n+2)(n+1)a_{n+2}x^n=\sum_{n=1}^\infty a_{n-1}x^n\]

Answer 7

yeh I usually have to write it out myself before I find an appropriate way, I believe starting at \(n=1\) is the simplest you can do here.

Answer 8

because the difference between n=1 and n=0 is usually adding a constant term.

Answer 9

You're very welcome.

Answer 10

Well actually I didn't change the \(C_{n+2}\) too, I just noticed that I made a mistake up there at my initial post, my bad *smiles*

Answer 11

I said the "dummy variable", it's how I was once told is called should be n+2, so I made the wrong substitution at the index for \(a_n\) obviously.

Answer 12

\[\large \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n= 2a_2 +6a_3x+12a_4x^2 +... \]
So you can interpret the RHS just as the initial value, meaning the sum evaluated at n=0 plus all that follows from that point, which is just the sum that starts by n=1 therefore you get:
\[ \large \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n= 2a_2 +\sum_{n=1}^\infty(n+2)(n+1)a_{n+2}x^{n}\] Notice that you can do that with most likely every starting point you desire, so I can also say that:
\[\large \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n= 2a_2 +6a_3x+\sum_{n=2}^\infty(n+2)(n+1)a_{n+2}x^n \]
Hope you see the pattern.

Answer 13

Because I like the exponential x^n, I think it's the one we could deal easily with. so we just have to worry about the runtime. I decided to go for n=1 because, as demonstrated above, that's usually the easiest one to compute. We could have tried to adjust the runtime of the RHS to go from n=0 too. but then we would have this ugly \(a_{-1}\) term in it, not really pleasant do deal with.

Answer 14

From what I remember for such kind of equations is that you always try to get the same exponential term, most people choose for x^n and the same run time (equal index values)

This is done so after that it's easier to compare the LHS to the RHS and then figure out the recursive equation. Usually it's a matter of taste what you solve for, but I believe this way is the most simple one.

Answer 15

I hope that helps :-) If not I will be back in a couple of minutes and check upon this, best of luck!