Use power series to solve the differential equation. $$y ^ { tt } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ {t } ( 0 ) = 0$$
A)$y ( x ) = 6 - \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }$
B)$y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }$
C)$y ( x ) = 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }$
D)$y ( x ) = - 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }$
E)$y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }$

Question

Quizplus · Accepted Answer

We assume that the solution is of the form:
$$y ( x ) = \sum _ { n = 0 } ^ { \infty } a _ { n } x ^ { n }$$
Differentiating twice with respect to $x$, we have
$$y ^ { t } ( x ) = \sum _ { n = 0 } ^ { \infty } a _ { n } n x ^ { n - 1 } , \qquad y ^ { tt } ( x ) = \sum _ { n = 0 } ^ { \infty } a _ { n } n ( n - 1 ) x ^ { n - 2 }$$
Substituting these expansions and the assumed series expansion of $y$ into the given differential equation: $$ \sum _ { n = 0 } ^ { \infty } a _ { n } n ( n - 1 ) x ^ { n - 2 } + \sum _ { n = 0 } ^ { \infty } x ^ { 2 } a _ { n } x ^ { n } = 0$$
Simplifying the above expression and solving for $a_0$ and $a_1$, we get:
$$ a_0=6, \qquad a_1=0 $$
Solving for $a_n$ for $n \geq 2$, we get
$$ a_{n} = \frac{(-1)^{n}6}{4!n(n-1)} a_{n-2} $$
Notice that $a_2$ is the first non-zero term. Substituting $n=2$ in the above recurrence relation and using $a_0=6$, we get
$$ a_2=-\frac{6}{2 \cdot 3 \cdot 4} $$
Again, substituting $n=4$, we get
$$ a_4=\frac{6}{4 \cdot 5 \cdot 8 \cdot 7} $$
Continuing in this fashion, we obtain
$$ a_{2k}=\frac{(-1)^k 6}{(4k)! (k!)}, \quad k \geq 1$$
Thus, the solution is given by:
$$y ( x ) = 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }$$

Use Power Series to Solve the Differential Equation ytt+x2y=0,y(0)=6,yt(0)=0y ^ { tt } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ {t } ( 0 ) = 0ytt+x2y=0,y(0)=6,yt(0)=0

Use Power Series to Solve the Differential Equation $y ^ { tt } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ {t } ( 0 ) = 0$