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Taylor's Theorem to Find the First Four Terms of the Series

Question 41

Multiple Choice

Taylor's Theorem to find the first four terms of the series solution of yt2xy=0y ^ {t } - 2 x y = 0 given the initial condition y(0) =1y ( 0 ) = 1 and use it to calculate y(3) y ( 3 ) . Round your answer to three decimal places wherever applicable.


A) y=1+22!x+124!x2+1206!x3+;y(3) 13.000y = 1 + \frac { 2 } { 2 ! } x + \frac { 12 } { 4 ! } x ^ { 2 } + \frac { 120 } { 6 ! } x ^ { 3 } + \cdots ; y ( 3 ) \approx 13.000
B) y=1+12!x2+64!x4+606!x6+,y(3) 86.500y = 1 + \frac { 1 } { 2 ! } x ^ { 2 } + \frac { 6 } { 4 ! } x ^ { 4 } + \frac { 60 } { 6 ! } x ^ { 6 } + \cdots , y ( 3 ) \approx 86.500
C) y=1+21!x2+122!x4+1203!x6+,y(3) 15,085.000y = 1 + \frac { 2 } { 1 ! } x ^ { 2 } + \frac { 12 } { 2 ! } x ^ { 4 } + \frac { 120 } { 3 ! } x ^ { 6 } + \cdots , y ( 3 ) \approx 15,085.000
D) y=1+22!x2+124!x4+1206!x6+;y(3) 172.000y = 1 + \frac { 2 } { 2 ! } x ^ { 2 } + \frac { 12 } { 4 ! } x ^ { 4 } + \frac { 120 } { 6 ! } x ^ { 6 } + \cdots ; y ( 3 ) \approx 172.000
E) y=1+21!x+122!x2+1203!x3+,y(3) 601.000y = 1 + \frac { 2 } { 1 ! } x + \frac { 12 } { 2 ! } x ^ { 2 } + \frac { 120 } { 3 ! } x ^ { 3 } + \cdots , y ( 3 ) \approx 601.000

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