What method to solve this difference equation?

Hi

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

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

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

I look forward to your responses!

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?

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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

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).

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?

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

Re: What method to solve this difference equation?

What about the factorial term? [Recall that g(n) = 5^n/n!]

Re: What method to solve this difference equation?

Oh yes!

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

How about now?

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]

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]

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

Thank you Chiro - I appreciate your patience and help. How about now?

Re: What method to solve this difference equation?

Re: What method to solve this difference equation?

Would this next step be correct?

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

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

Re: What method to solve this difference equation?

First solve the homogeneous equation

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

Work it out from

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

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

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

etc.

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

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

That gets you the solution $\displaystyle u_{n}=(2+n)\frac{5^{n}}{n!}.$

Re: What method to solve this difference equation?

Thank you so so much! I wish you were my lecturer!!

Re: What method to solve this difference equation?