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

question 13

Multiple Choice

Use Taylor's Theorem to find the first eight terms of the series solution of $y ^ { tt } - 2 x y ^ { t } + y = 0$ given the initial conditions $y ( 0 ) = 1 , y ^ { t} ( 0 ) = 4$ and use it to calculate $y \left( \frac { 1 } { 3 } \right)$ .
Round your answer to three decimal places.

• A) \(y = 1 + \frac { 4 } { 1 ! } x - \frac { 1 } { 2 ! } x ^ { 2 } + \frac { 4 } { 3 ! } x ^ { 3 } + \frac { 3 } { 4 ! } x ^ { 4 } + \frac { 20 } { 5 ! } x ^ { 5 } + \frac { 21 } { 6 ! } x ^ { 6 } + \frac { 180 } { 7 ! } x ^ { 7 } + \cdots ; y \left( \frac { 1 } { 3 } \right) \leqslant 2.305\)
• B) \(y = 1 + \frac { 4 } { 1 ! } x - \frac { 1 } { 2 ! } x ^ { 2 } + \frac { 4 } { 3 ! } x ^ { 3 } - \frac { 3 } { 4 ! } x ^ { 4 } + \frac { 20 } { 5 ! } x ^ { 5 } - \frac { 21 } { 6 ! } x ^ { 6 } + \frac { 180 } { 7 ! } x ^ { 7 } + \cdots ; y \left( \frac { 1 } { 3 } \right) \approx 2.302\)
• C)\(y = 1 - \frac { 4 } { 1 ! } x + \frac { 2 } { 2 ! } x ^ { 2 } - \frac { 4 } { 3 ! } x ^ { 3 } + \frac { 3 } { 4 ! } x ^ { 4 } - \frac { 4 } { 5 ! } x ^ { 5 } + \frac { 4 } { 6 ! } x ^ { 6 } - \frac { 4 } { 7 ! } x ^ { 7 } + \cdots ; y \left( \frac { 1 } { 3 } \right) \approx - 0.468\)
• D)
• \(y = 1 + \frac { 4 } { 1 ! } x + \frac { 1 } { 2 ! } x ^ { 2 } + \frac { 12 } { 3 ! } x ^ { 3 } + \frac { 3 } { 4 ! } x ^ { 4 } + \frac { 28 } { 5 ! } x ^ { 5 } - \frac { 84 } { 6 ! } x ^ { 6 } + \frac { 45 } { 7 ! } x ^ { 7 } + \cdots , y \left( \frac { 1 } { 3 } \right) \approx 2.465\)
• E)
• \(y = 1 + \frac { 4 } { 1 ! } x - \frac { 1 } { 2 ! } x ^ { 2 } + \frac { 12 } { 3 ! } x ^ { 3 } - \frac { 3 } { 4 ! } x ^ { 4 } + \frac { 28 } { 5 ! } x ^ { 5 } - \frac { 84 } { 6 ! } x ^ { 6 } + \frac { 45 } { 7 ! } x ^ { 7 } + \cdots , y \left( \frac { 1 } { 3 } \right) \approx 2.351\)

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