Use power series to solve the differential equation.
$$y ^ { \prime \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { t } ( 0 ) = 0$$
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

To solve the given differential equation using power series, we assume that the solution can be expressed as
$$y(x) = \sum_{n=0}^{\infty} a_n x^n$$
Taking derivatives, we get
$$y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}$$
$$y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$
Substituting these into the differential equation and simplifying, we get
$$\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + \sum_{n=0}^{\infty} x^2 a_n x^n = 0$$
Since this equation must hold for all values of $x$, we can match coefficients of powers of $x$. The coefficient of $x^0$ gives us $2a_2 = 0$ or $a_2 = 0$ since $a_2$ cannot be zero. The coefficient of $x^1$ gives us $6a_3=0$ or $a_3=0$ since $a_3$ cannot be zero. The coefficient of $x^n$ for $n\geq 2$ gives us
$$n(n-1)a_n + a_{n-2} = 0$$
which can be rearranged as
$$a_n = -\frac{a_{n-2}}{n(n-1)}$$
Using the initial conditions $y(0) = 6$ and $y'(0) = 0$, we can determine the values of $a_0$ and $a_1$ as
$$a_0 = 6$$
$$a_1 = 0$$
Substituting these values and the recurrence relation for $a_n$ into the power series formula for $y$, we get
$$y(x) = 6 + \sum_{n=2}^{\infty} (-1)^n \frac{x^{4n}}{4n(4n-1)(4n-4)\ldots 4\cdot 3}$$
which is equivalent to choice C.

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

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