Solve the difference equation?

In summary: You should start again, and this time, write down what yn+1 is supposed to be, and how it is related to yn.But what you wrote had nothing to do with solving a differential equation either. You should start again, and this time, write down what yn+1 is supposed to be, and how it is related to yn.
  • #1
Success
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0
Solve the difference equation yn+1=(n+1)/(n+2) yn in terms of the initial value y0.
 
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  • #2
Hi Success! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

(common-sense should solve this)
 
  • #3
I really don't even know how to start. I guess you should begin with yn=1/(n+1) y0.
 
  • #4
Use Quotient Rule

[itex]\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{vu'-uv'}{v^2}[/itex]

You should be able to solve this.
 
  • #5
Success said:
I really don't even know how to start. I guess you should begin with yn=1/(n+1) y0.

do you mean y1 = 1/(2) y0 ?

yes you could start with that, then find y2, then y3, …

and see if you can spot a pattern :smile:
 
  • #6
Success said:
I really don't even know how to start. I guess you should begin with yn=1/(n+1) y0.
IF that is true then it is the answer to your problem! If you can start writing down the answer, you surely don't need our help! How did you get that?

Poor, not so brilliant people like me might start by writing out a few values and looking for a pattern. If [itex]y_{n+1}= ((n+1)/(n+2))y_n[/itex], then [itex]y_1= ((0+1)/(0+2))y_0= y_0/2[/itex], [itex]y_2= ((1+1)/(1+2))y_1= 2y_1/3= 2(y_0/2)/3= y_0/3[/itex], [itex]y_3= ((2+1)/(2+2)y_2= (3/4)y_2= (3/4)(y_0/3)= y_0/4[/itex], [itex]y_4= ((3+1)/(3+2)y_3= (4/5)y_3= (4/5)(y_0/4)= y_0/5[/itex]...

Do you think you can make a guess now? To be complete, you should then prove that your guess is correct, by induction, say.
 
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  • #7
jhosamelly, I got 3/(n+2)^2 by quotient rule. How is that the answer?
 
  • #8
Thanks everyone.
 
  • #9
Success said:
jhosamelly, I got 3/(n+2)^2 by quotient rule. How is that the answer?
This problem has nothing to do with Calculus, the derivative, or the quotient rule.
 
  • #10
HallsofIvy said:
This problem has nothing to do with Calculus, the derivative, or the quotient rule.

Ow, sorry, I thought your title is "solve the DIFFERENTIAL equation" .
 
  • #11
spot the difference! :biggrin:
 
  • #12
jhosamelly said:
Ow, sorry, I thought your title is "solve the DIFFERENTIAL equation" .
But what you wrote had nothing to do with solving a differential equation either.
 

1. What is a difference equation?

A difference equation is a mathematical equation that describes how a variable changes over time. It is used to model systems that evolve over discrete time steps, rather than continuously.

2. How do you solve a difference equation?

The process for solving a difference equation involves finding a general solution that satisfies the equation, and then using initial conditions to determine specific values for the variables. This can be done using various techniques, such as substitution, iteration, or using generating functions.

3. What is the difference between a difference equation and a differential equation?

While both types of equations describe how a variable changes over time, the main difference is that a difference equation uses discrete time steps, while a differential equation uses continuous time. This means that a difference equation can only be solved for specific values, while a differential equation can be solved for any value within a given range.

4. What are some real-world applications of difference equations?

Difference equations have many applications in various fields, such as economics, physics, and engineering. They are often used to model population growth, chemical reactions, and electrical circuits, among other things.

5. Can difference equations be solved analytically or numerically?

Difference equations can be solved using both analytical and numerical methods. Analytical solutions involve finding a general formula for the solution, while numerical solutions involve using algorithms to approximate the solution at specific time steps. The method used will depend on the complexity of the equation and the desired level of accuracy.

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