Question 5: (1 point) In order to keep the songbirds in the back yard happy, Jür

Question

Question 5: (1 point) In order to keep the songbirds in the back yard happy, Jürgen puts out 25 g of seeds at the end of each week. During the week, the birds find and eat 415 of the available seeds. The DTDS for the amount of seeds in the back yard is +S)S +25, where is measured in weeks and seeds are counted just after Jürgen brings in the new supply. a) What is the updating function of the DTDS? Answer:f(S) b) Find the fixed point (also called equilibrium point) S* of this linear DTDS. Give an exact answer. Answer: S c) Give the general solution of this linear DTDS, as a function of both I and the initial condition S. (Use S to denote So because Maple TA doesn't understand subscripts.) Answer: S

Explanation / Answer

a) f(s) = (1/5)s+25

b) For equilibrium point we have st+1 = st.

Then, st = (1/5)st+25

i.e., st-(1/5)st = 25

i.e., [1-(1/5)]st = 25

i.e., (4/5)st = 25

i.e., st = 25*5/4

i.e., st = 125/4

Therefore, the fixed point of the linear DTDS is : S* = st ,i.e., S* = 125/4.

c) Given, st+1 = (1/5)st+25

i.e., st+1 = (1/5)[(1/5)st-1+25]+25

i.e., st+1 = (1/5)2st-1+5+25

i.e., st+1 = (1/5)3st-2+1+5+25

Proceeding in this way we get, st+1 = (1/5)t+1s+[25+5+1+(1/5)+.....+(1/5)t-2]

Then, st = (1/5)ts+[25+5+1+(1/5)+.....+(1/5)t-3]

i.e., st = (1/5)ts+25*[1+(1/5)+(1/5)2+.......+(1/5)t-1]

i.e., st = (1/5)ts+25*[1-(1/5)t]/[1-(1/5)]

i.e., st = (1/5)ts+25*[1-(1/5)t]/(4/5)

i.e., st = (1/5)ts+(25*5)*[1-(1/5)t]/4

i.e., st = (1/5)ts+125*[1-(1/5)t]/4

Therefore, the general solution for the linear DTDS is : st = (1/5)ts+125*[1-(1/5)t]/4.

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