# What method to solve this difference equation?

• Dec 23rd 2012, 08:45 AM
zzizi
What method to solve this difference equation?
Hi

Could someone shed some light, on how I would solve the following difference equation?

$u_{n}=\frac{5}{n}u_{n-1} + \frac{5^{n}}{n!}, u_{0}=2$

Would I use the divide and conquer method or something else?

I look forward to your responses!
• Dec 23rd 2012, 01:22 PM
chiro
Re: What method to solve this difference equation?
Hey zzizi.

For these kinds of problems, you should consider getting a_n as a function of n purely and see if there is a relationship.

Also do you have a set of identities or formulas for your class that you are doing this exercise for?
• Dec 23rd 2012, 02:02 PM
zzizi
Re: What method to solve this difference equation?
Quote:

Originally Posted by chiro
Hey zzizi.

For these kinds of problems, you should consider getting a_n as a function of n purely and see if there is a relationship.

Also do you have a set of identities or formulas for your class that you are doing this exercise for?

Hi Chiro!

Yes I have been given a formula - I have attached it. Please have a look, I'd appreciate help in trying to apply this problem using the given formula. Thank youAttachment 26336
• Dec 23rd 2012, 02:11 PM
chiro
Re: What method to solve this difference equation?
Can you identify what the f's and g's should be in this formula? (It's mostly a thing of plugging in the definitions).
• Dec 23rd 2012, 02:22 PM
zzizi
Re: What method to solve this difference equation?
Quote:

Originally Posted by chiro
Can you identify what the f's and g's should be in this formula? (It's mostly a thing of plugging in the definitions).

Is this correct?

$2\prod_{i=1}^{n}\frac{5}{i}+\sum_{i=1}^{n}\frac{5^ {i}}{i}.\prod_{j=i+1}^{n}\frac{5}{j}$
• Dec 23rd 2012, 03:49 PM
chiro
Re: What method to solve this difference equation?
What about the factorial term? [Recall that g(n) = 5^n/n!]
• Dec 23rd 2012, 03:54 PM
zzizi
Re: What method to solve this difference equation?
Oh yes!

$2\prod_{i=1}^{n}\frac{5}{i}+\sum_{i=1}^{n}\frac{5^ {i}}{i!}.\prod_{j=i+1}^{n}\frac{5}{j!}$

• Dec 23rd 2012, 03:56 PM
chiro
Re: What method to solve this difference equation?
Recall that f(n) = 5/n [Hint: Look at the last multiplication term, you've confused f(n) and g(n) using the definition]
• Dec 23rd 2012, 04:00 PM
zzizi
Re: What method to solve this difference equation?
Quote:

Originally Posted by chiro
Recall that f(n) = 5/n [Hint: Look at the last multiplication term, you've confused f(n) and g(n) using the definition]

$2\prod_{i=1}^{n}\frac{5}{i}+\sum_{i=1}^{n}\frac{5^ {i}}{i!}.\prod_{j=i+1}^{n}\frac{5}{j}$

Thank you Chiro - I appreciate your patience and help. How about now?
• Dec 23rd 2012, 05:22 PM
chiro
Re: What method to solve this difference equation?
That looks pretty good.
• Dec 24th 2012, 10:33 AM
zzizi
Re: What method to solve this difference equation?
Would this next step be correct?

$2.\frac{5^{n}}{n!}+\sum_{i=1}^{n}\frac{5^{i}}{i!}. \frac{5^{n}i!}{n!5^{i}}$
• Dec 25th 2012, 09:43 AM
zzizi
Re: What method to solve this difference equation?
Can anyone help me with this next step? or possibly offer a link to site that offers a tutorial on this subject. Is this called the telescoping method? Many thanks in advance
• Dec 25th 2012, 03:37 PM
BobP
Re: What method to solve this difference equation?
First solve the homogeneous equation

$w_{n}=\frac{5}{n}w_{n-1}.$

Work it out from

$w_{1}= \frac{5}{1}w_{0},$

$w_{2}=\frac{5}{2}w_{1}=\frac{5^{2}}{2!}w_{0},$

$w_{3}=\frac{5}{3}w_{2}=\frac{5^{3}}{3!}w_{0},$

etc.

Having done that, assume that the solution to the original equation is $u_{n}=v_{n}w_{n}.$

Substitute that in and show that $v_{n}=1+\frac{n}{2}.$

That gets you the solution $u_{n}=(2+n)\frac{5^{n}}{n!}.$
• Dec 25th 2012, 03:49 PM
zzizi
Re: What method to solve this difference equation?
Thank you so so much! I wish you were my lecturer!!
• Dec 25th 2012, 04:42 PM
zzizi
Re: What method to solve this difference equation?
.