Solving a Linear ODE using a power series

[yecko]

Homework Statement

Homework Equations

Power series
ODE

The Attempt at a Solution

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Sorry for not typing all those things out from my phone..
How can I get C1?
And how can I put the solution in the required format? (I don't know how to put it in summation sign... and i cannot even solve out for r in the question...)
Thank you so much for any help!

 

[BvU]

cannot even solve out for r in the question

Small wonder: you leave out $r$ from the derivatives of $y$. Except in $C_n$, you want to replace $n$ by $n+r$

 

[yecko]

Thank you.
However even i used r+n, i still cannot make it into the required form... did i have something wrong with it? Or what should I do next to obtain the answer in the required format? Thank you.

 

[ehild]

You have a mistake in the equation for r. It should be r²+4r+4=0

 

[yecko]

Thanks for pointing this out...

r=-2 which is now correct,
However, i cannot solve out the denominator of the summation part... how can I write it in form of summation?
Thank you very much!

 

[ehild]

Substitute r = -2 and simplify the denominator of a_n.

 

[yecko]

I have already substituted r=-2, but i don't know the recurrance relation of that denominator... do you mind to give me some hints or relationship related?
Thank you

 

[ehild]

Type in the recurrence relation of a_n, please. Your handwriting is blurred,

I have already substituted r=-2, but i don't know the recurrance relation of that denominator... do you mind to give me some hints or relationship related?
Thank you

You got the recurrence relation for a_n:
$$a_n=\frac{4 a_{n-1}}{(n+r)(n+r-1)+5(r+n)+4}$$
Substitute r=-2 and simplify. What do you get?

 

[yecko]

Sorry for blur image... let me post a clearer image here:

I have got my answer as 4^n/(n!), yet it is wrong as by the submission system... what's wrong with my answer? Thank you

 

[yecko]

 

[ehild]

Sorry for blur image...
View attachment 223115
I have got my answer as 4^n/(n!), yet it is wrong as by the submission system... what's wrong with my answer? Thank you

The recurrence relation for a_n is

$a_n=\frac {4 a_{n-1}}{n^2}$, with a₀=1.
How did you get n! in the numerator?

 

[yecko]

As I have to write the answer in terms of n only as a summation form (refer to the question image I have posted in #1), only writting it in recurrance relationship is not enough...
As the image I posted in #9 post explained how i got “n!”, when n tends larger, the relationship a(n) tends to be n! in denominator

4^n*(n!)/(n!)^2 =4^n/(n!)

 

[yecko]

I have already tried my best in obtaining the answer, do you mind to lead me what should I write in the blank as of the question image posted in post #10 please?
Thank you so much for your help!

 

[ehild]

I have already tried my best in obtaining the answer, do you mind to lead me what should I write in the blank as of the question image posted in post #10 please?
Thank you so much for your help!

I do not understand what you mean on "summation form"
You need to find the coefficients c_n in the power series solution.
You derived the recurrence relation $$c_n=\frac{4}{n^2}c_{n-1}$$, but you wrote wrong results for c₁, c₂, c₃, c₄...
c₀=1
$$c_1=\frac{4}{1^2}c_{0}= \frac{4}{1^2}$$
$$c_2=\frac{4}{2^2}c_{1}=\frac{4^2}{2^2\cdot1^2}=\frac{4^2}{(2!)^2}$$
$$c_3=\frac{4}{3^2}c_{2}=??$$
and so on. So what is c_n?

 

[yecko]

Oh! I get it! I made a mistake in the power of denominator! Thanks for reminding!

(Just an ad, can you also help me with a similar question in thread https://www.physicsforums.com/threa...uation-with-power-series.943404/#post-5969575 ? Thanks in advance：)))))