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

question 8

Multiple Choice

Use Taylor's Theorem to find the first four terms of the series solution of $y ^ { tt } + e ^ { x } y ^ {t } - ( \sin x ) y = 0$ given the initial conditions $y ( 0 ) = - 5$ , and $y ^ { t } ( 0 ) = 7$ and use it to calculate $y \left( \frac { 1 } { 4 } \right)$ . Round your answer to three decimal places.

• A) \(y = - 5 + \frac { 7 } { 1 ! } x - \frac { 5 } { 2 ! } x ^ { 2 } - \frac { 7 } { 3 ! } x ^ { 3 } + \cdots , y \left( \frac { 1 } { 4 } \right) \approx - 3.424\)
• B) \(y = - 5 + \frac { 7 } { 1 ! } x - \frac { 5 } { 2 ! } x ^ { 2 } + \frac { 7 } { 3 ! } x ^ { 3 } + \cdots ; y \left( \frac { 1 } { 4 } \right) \approx - 3.388\)
• C) \(y = - 5 + \frac { 7 } { 1 ! } x - \frac { 21 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \cdots , y \left( \frac { 1 } { 4 } \right) \approx - 3.932\)
• D)\(y = - 5 + \frac { 7 } { 1 ! } x - \frac { 7 } { 2 ! } x ^ { 2 } + \frac { 5 } { 3 ! } x ^ { 3 } + \cdots , y \left( \frac { 1 } { 4 } \right) \approx - 3.456\)
• E)\(y = - 5 + \frac { 7 } { 1 ! } x - \frac { 7 } { 2 ! } x ^ { 2 } - \frac { 5 } { 3 ! } x ^ { 3 } + \cdots ; y \left( \frac { 1 } { 4 } \right) \approx - 3.482\)

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