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)
$y = 1 + \frac { 4 } { 1 ! } x - \frac { 1 } { 2 ! } x ^ { 2 } + \frac { 12 } { 3 ! } x ^ { 3 } - \frac { 3 } { 4 ! } x ^ { 4 } + \frac { 28 } { 5 ! } x ^ { 5 } - \frac { 84 } { 6 ! } x ^ { 6 } + \frac { 45 } { 7 ! } x ^ { 7 } + \cdots , y \left( \frac { 1 } { 3 } \right) \approx 2.351$

Question

Quizplus · Accepted Answer

Using Taylor's Theorem, we have
$$y=x+\frac{y(0)}{1!}x^2+\frac{y'(0)}{2!}x^3+\frac{y''(0)}{3!}x^4+\cdots$$
Differentiating the given differential equation gives $y^{(4)}-2xy''=0$, which, after simplification, gives $y''''=6y'''-12y''$. Similarly, we can find $y^{(6)}$ and $y^{(8)}$ and substitute the initial conditions to get the first eight terms of the series as:
$$y=1+4x-\frac{1}{2}x^2+\frac{4}{3!}x^3-\frac{3}{4!}x^4+\frac{20}{5!}x^5-\frac{21}{6!}x^6+\frac{180}{7!}x^7+\cdots$$
To find $y\left(\frac13\right)$, we substitute $x=\frac13$ in the above expression and round off to three decimal places to get $2.302$. Thus, the answer is option B.

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