Use power series to solve the differential equation. Select the correct answer.
$$y ^ { tt } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { t } ( 0 ) = 0$$
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 can use the power series method to solve the given differential equation. Assume that the solution can be expressed as a power series:
$$y(x) = \sum_{n=0}^\infty a_n x^n$$
Differentiating twice, we get:
$$y''(x) = \sum_{n=2}^\infty n(n-1) a_n x^{n-2}$$
Multiplying the given differential equation by $x^2$ and substituting $y$ and $y''$ as power series, we get:
$$\sum_{n=2}^\infty n(n-1) a_n x^n + \sum_{n=0}^\infty a_n x^{n+2} = 0$$
Shifting the index of the second summation, we get:
$$\sum_{n=2}^\infty n(n-1) a_n x^n + \sum_{n=2}^\infty a_{n-2} x^n = 0$$
Comparing the coefficients of the same powers of $x$, we get the recurrence relation:
$$a_n = -\frac{a_{n-2}}{n(n-1)} \quad 	ext{for } n \geq 2$$
Using the initial conditions $y(0)=6$ and $y'(0)=0$, we get:
$$a_0 = 6 \quad 	ext{and} \quad a_1 = 0$$
Substituting the recurrence relation repeatedly, we can express all the coefficients in terms of $a_0$ and $a_1$, which gives us the power series solution of the differential equation. Simplifying the expression obtained by substituting the coefficients, we get the choice C as the correct answer.

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 $y ^ { tt } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { t } ( 0 ) = 0$