Solve the difference equation?

[Success]

Solve the difference equation y_n+1=sqrt((n+3)/(n+1)) y_n in terms of the initial value y₀.

y₁=sqrt(3)y₀
y₂=sqrt(6)y₀
y_n=sqrt(3n)y₀

But that's not the answer. The answer in the textbook is y_n=y₀sqrt(((n+2)(n+1))/2). Did I do something wrong?

 

[HallsofIvy]

It is hard to say what you might have done wrong when you do not show what you did!

While it is a good idea to calculate values until you can guess a pattern, you are not done until you have proved that pattern. At the very least, you should have checked your "guess". [itex]y_3= \sqrt{\frac{2+3}{2+ 1}}\sqrt{6}y_0= \sqrt{\frac{5(6)}{3}}y_0= \sqrt{10}y_0[/itex]. 10 is NOT 3n= 3(3)= 9. Certainly, you should have realized that "[itex]y_n= \sqrt{3n}y_0[/itex]" is not true when n= 0!

Yes, [itex]y_1= \sqrt{3}y_0[/itex] and [itex]y_2= \sqrt{6}y_0[/itex]. But then [itex]y_3= \sqrt{10}y_0[/itex], as I showed, and then [itex]y_4= \sqrt{(3+3)/(3+ 1)}\sqrt{10}y_0= \sqrt{15}y_0[/itex], etc.6,

Is that enough to make a better guess at a pattern? I notice that it is always the square root of an integer times [itex]y_0[/itex]. Further, those integers are 3, 6, 10, 15... For me, at least, that is enough to notice that the 'first differences' are 6-3= 3, 10- 6= 4, 15- 10= 5 and then the 'second differences' are 4- 3= 1 and 5- 4= 1- both 1. IF that continued to be true, that the 'second difference' is a constant, that would mean that the integers are a quadratic function of n, [itex]an^2+ bn+ c[/itex]and we can use those values to determine a, b, and c.
I did that and got that a= 1/2, b= 3/2, and c= 1 so that the integers are [itex](1/2)(n^2+ 3n+ 2)= (1/2)(n+ 2)(n+ 1)[/itex], exactly what your textbook says.

You can prove that is true using "proof by induction". When n= 0, [itex]y_0= \sqrt{\frac{1}{2}(0+2)(0+ 1)}y_0= \sqrt{1}y_0= y_0[/itex] is correct. Assume that, for some k, [itex]y_k= \sqrt{\frac{1}{2}(k+2)(k+ 1)}y_0[/itex]. Then, by the difference equation, [itex]y_{k+1}= \sqrt{(k+3)/(k+1)}y_k= \sqrt{\frac{1}{2}((k+3)/(k+1))(k+2)(k+1)}y_0[/itex][itex]= \sqrt{\frac{1}{2}(k+3)(k+2)}y_0= \sqrt{\frac{1}{2}((k+1)+ 2)((k+1)+ 1)}y_0[/itex].