2nd order differential equation with power series

[yecko]

Homework Statement

Homework Equations

Power series

The Attempt at a Solution

As I have to write in form of "x^2n" & "x^2n+1", I am totally have no idea with how can I go on to do the question.
Those I have learned in lecture and online are mostly with only one part of summation... or two parts with two distinctive roots and two constants with it...
Can anyone guide me with what I should do next in order to obtain the answer please? Thank you.

 

[Ray Vickson]

Homework Statement

View attachment 222977

Homework Equations

Power series

The Attempt at a Solution

View attachment 222976
As I have to write in form of "x^2n" & "x^2n+1", I am totally have no idea with how can I go on to do the question.
Those I have learned in lecture and online are mostly with only one part of summation... or two parts with two distinctive roots and two constants with it...
Can anyone guide me with what I should do next in order to obtain the answer please? Thank you.

Your posted image is perfectly clear and derives the result more-or-less completely. I can see no way to explain it better than what you are already seeing.

However, I will offer a little hint: If $y = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots + c_n x^n + \cdots$ then you can write down the series for the left-hand-side of the DE---just do the derivatives and add everything together. The result will have the form $y'' -5 x y'-5y = D_0 + D_1 x + D_2 x^2 + \cdots ,$ where $D_0, D_1, D_2, \ldots$ are combinations of $c_0, c_1, c_2, \ldots$. Just do some algebra to figure out how the $D_i$ and $c_j$ are related.

However, the result should be $0$ (because $y$ is supposed to solve the DE), so you should have $D_0 = 0, D_1 = 0, D_2 = 0, \cdots .$ That will give you a bunch of equations that relate the different $c_j$ to each other. The two initial conditions give you the rest of the needed information, so you can find all the $c_j$.

If you cannot see right away how to do the general case (that is, $D_m$ for general, symbolic $m$) start with writing down all the terms for $D_m$ for $m = 0,1,2, \ldots, 10,$ say, and see what these 11 equations give you. There are no shortcuts---you need to sit down and work it all out.

 

[yecko]

This is the subsequent steps I have done like what you mentioned, and I have found the relationship. However, I don’t know how to write it in terms of summation sign... how can I write my answer in order to fulfill the format of the question?
Thank you

 

[vela]

The recurrence relation you derived for $c_n$ is incorrect. You dropped a factor of 5 on the last term.

Whenever you have something multiplied by 0, you can just set it to 0. There's no sense in deriving a general expression for the odd-powered terms when you know the coefficients are 0 because $c_1=0$ to satisfy the initial conditions.

 

[yecko]

The recurrence relation you derived for $c_n$ is incorrect. You dropped a factor of 5 on the last term.

Whenever you have something multiplied by 0, you can just set it to 0. There's no sense in deriving a general expression for the odd-powered terms when you know the coefficients are 0 because $c_1=0$ to satisfy the initial conditions.

Thanks for reminding.
however, I am still stuck with even power terms,
with a(2n)={[5(2n+1)]*[5(2n-1)]*[5(2n-3)]*...*[5(1)]}/(2n)!=(5^n)*{(2n+1)*(2n-1)*(2n-3)*...*1}/(2n)!
how to make it into summation sign?
thank you

 

[vela]

Try simplifying the recurrence relation before trying to work out what $a_{2n}$ looks like.

 

[yecko]

(5^n)*{(2n+1)*(2n-1)*(2n-3)*...*1}/(2n)!
This is the simpliest form i can think of with a(2n)
Have i made any mistake so that i cannot simplify it?
Can you give me some hints on how to simplify the part {(2n+1)*(2n-1)*(2n-3)*...*1} ?
Thank you

 

[vela]

(5^n)*{(2n+1)*(2n-1)*(2n-3)*...*1}/(2n)!
This is the simpliest form i can think of with a(2n)

That's not the recurrence relation, is it? If you make the obvious simplification, you'll get a slightly simpler expression for $a_{2n}$.

Have i made any mistake so that i cannot simplify it?
Can you give me some hints on how to simplify the part {(2n+1)*(2n-1)*(2n-3)*...*1} ?
Thank you

Alternatively, you can use your current expression and write out $a_2, a_4, a_6, \dots$. There should be a pretty obvious pattern of cancellations.

 

[yecko]

This is what I calculate this time:

However, the system regards me wrong about my domain... have I calculated anything wrong with it?

Thank you

 

[vela]

Consider the $n=1$ term in the first summation. The coefficient in your sum would be $a_1 = \frac{5^{1/2}}{2(1/2)!}.$

 

[yecko]

So what should I do in order to avoid that?
I have tried 5^n/(2(n!)) which is also wrong