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Use Taylor's Theorem to Find the First Six Terms of the Series

Question 30

Multiple Choice

Use Taylor's Theorem to find the first six terms of the series solution of  of ytt2xy=0\text { of } y ^ { tt } - 2 x y = 0 given the initial conditions y(0) =5y ( 0 ) = 5 and yt(0) =7y ^ {t } ( 0 ) = - 7


A)
y=571!x+103!x3284!x4+806!x62807!x7+y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 3 ! } x ^ { 3 } - \frac { 28 } { 4 ! } x ^ { 4 } + \frac { 80 } { 6 ! } x ^ { 6 } - \frac { 280 } { 7 ! } x ^ { 7 } + \cdots
B) y=751!x+143!x3204!x4+1126!x62007!x7+y = 7 - \frac { 5 } { 1 ! } x + \frac { 14 } { 3 ! } x ^ { 3 } - \frac { 20 } { 4 ! } x ^ { 4 } + \frac { 112 } { 6 ! } x ^ { 6 } - \frac { 200 } { 7 ! } x ^ { 7 } + \cdots
C) y=751!x+143!x3204!x4+1127!x71758!x8+y = 7 - \frac { 5 } { 1 ! } x + \frac { 14 } { 3 ! } x ^ { 3 } - \frac { 20 } { 4 ! } x ^ { 4 } + \frac { 112 } { 7 ! } x ^ { 7 } - \frac { 175 } { 8 ! } x ^ { 8 } + \cdots
D) y=571!x+103!x3284!x4+907!x72458!x8+y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 3 ! } x ^ { 3 } - \frac { 28 } { 4 ! } x ^ { 4 } + \frac { 90 } { 7 ! } x ^ { 7 } - \frac { 245 } { 8 ! } x ^ { 8 } + \cdots
E) y=571!x+102!x2283!x3+804!x42805!x5+y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 2 ! } x ^ { 2 } - \frac { 28 } { 3 ! } x ^ { 3 } + \frac { 80 } { 4 ! } x ^ { 4 } - \frac { 280 } { 5 ! } x ^ { 5 } + \cdots

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