Use power series to solve the differential equation. $y ^ { \prime \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { \prime } ( 0 ) = 0$
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 first write the power series for $y(x)$ as $$y(x)=\sum_{n=0}^\infty a_{n}x^n$$ where $a_{n}=\dfrac{1}{n!}y^{(n)}(0)$.

Since $y(0)=6$, we have $a_0=6$. Also, we know that $y'(0)=0$, so $a_1=0$.

To find the rest of the coefficients, we differentiate the given differential equation with respect to $x$ repeatedly and substitute $x=0$. The differential equation gives us $$y''(x)=-x^2y(x)$$ Differentiating both sides of this equation with respect to $x$, we get $$y'''(x)=-2xy'(x)-x^2y''(x)$$ Substituting $x=0$ in this equation, we get $$a_3=-\dfrac{1}{3\cdot 2}a_0=-1$$
Differentiating both sides of this equation with respect to $x$ again, we get $$y^{(4)}(x)=-4y''(x)-x^2y'''(x)$$ Substituting $x=0$ in this equation, we get $$a_4=-\dfrac{1}{4\cdot 3}a_2=0$$ where $a_2=-a_0/2$ is the coefficient of $x^2$ in $y(x)$. We can use this recursion formula in general to find all $a_{2n}$:$$a_{2n}=-\dfrac{1}{(2n)(2n-1)}a_{2n-2}$$

Similarly, for $y^{(2n+1)}(0)=0$, we get $a_{2n+1}=0$. Thus, we have $$y(x)=a_0+a_2x^2+a_4x^4+\cdots =6+\sum_{n=1}^\infty (-1)^n\dfrac{6x^{4n}}{(4n)!/4}$$
where we have used the recursion formula to get the coefficient of $x^{4n}$.

Simplifying this expression, we get 
$$y(x)=6+\sum_{n=1}^\infty (-1)^n\dfrac{6x^{4n}}{4n(4n-1)(4n-2)(4n-3)}$$
which matches choice (C).

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

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