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Solve the difference equation?

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

y1=sqrt(3)y0

y2=sqrt(6)y0

y3=sqrt(10)y0

Answer: yn=y0*sqrt(((n+2)(n+1))/2)

How do I get to the answer?

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1 Answer

Substitute n for n+1,

yn

= sqrt((n+2)/(n))yn-1

= sqrt((n+2)(n+1)/(n(n-1)))yn-2

= sqrt((n+2)(n+1)n/(n(n-1)(n-2)))yn-3

= sqrt((n+2)(n+1)n...3/(n(n-1)(n-2)...1))y0

= y0*sqrt(((n+2)(n+1))/2), since (n+2)(n+1)n...3 = (n+2)!/2 = (n+2)(n+1)n!/2

 

QED