Success
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Solve the difference equation yn+1=(n+1)/(n+2) yn in terms of the initial value y0.
How to Solve the Difference Equation yn+1=(n+1)/(n+2) yn?
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Hi Success! 
Show us what you've tried, and where you're stuck, and then we'll know how to help!
(common-sense should solve this)
Show us what you've tried, and where you're stuck, and then we'll know how to help!
(common-sense should solve this)
Success
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I really don't even know how to start. I guess you should begin with yn=1/(n+1) y0.
jhosamelly
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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.
[itex]\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{vu'-uv'}{v^2}[/itex]
You should be able to solve this.
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Success said:
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
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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?Success said:
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.
Success
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jhosamelly, I got 3/(n+2)^2 by quotient rule. How is that the answer?
Success
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Thanks everyone.
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This problem has nothing to do with Calculus, the derivative, or the quotient rule.Success said:
jhosamelly
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HallsofIvy said:
Ow, sorry, I thought your title is "solve the DIFFERENTIAL equation" .
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spot the difference! 
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But what you wrote had nothing to do with solving a differential equation either.jhosamelly said: