Solve the given difference equation in terms of the initial condition $y_0$. Describe the behavior of the solution as $n\rightarrow\infty$

$\large{y_{n+1}=\sqrt{\dfrac{n+3}{n+1}}y_n}$

Question

Solve the given difference equation in terms of the initial condition $y_0$. Describe the behavior of the solution as $n\rightarrow\infty$

$\large{y_{n+1}=\sqrt{\dfrac{n+3}{n+1}}y_n}$

austinl · Answer

Um.... what?? I am sorry, missed the lecture for this so I am completely lost.

atlas · Answer

$$y _{n+1}^2 = \frac{ (n+3)}{( n+1) } y_n^2$$

atlas · Answer

y1^2/y0^2 = 3 .....(i)
y2^2/y1^2 = 2 ......(ii)
..........
yn+1^2/yn^2 = n+3/n+1 ....
Multiply all of them

atlas · Answer

$$\frac{ y ^2_{n+1} }{ y ^2_{0} } = \Pi(n+3)/(n+1) $$

atlas · Answer

the limits of pi are from n=1 to n+1

atlas · Answer

did you get the answer ??

austinl · Answer

I'm trying to figure out what you just did...

atlas · Answer

i m off to sleep .......the answer i guess is (n+2)(n+3) /2.............

austinl · Answer

Ok, goodnight and thanks for the help!

atlas · Answer

you are welcome......btw I never took a difference equation class but the trick everytime is to cancel out all other yn's by subtraction or division