Kansas State University

STEP 1: Plug in $\displaystyle y(x) = \sum_{n=0}^{\infty}a_n x^n $ and compute all the different terms in the equation $$ \begin{align} 8y &= \sum_{ n = 0 }^{ \infty }\,8 a_{ n } x^{ n }\\ (-x^2 - 8)y' &= \sum_{ n = 1 }^{ \infty }\,-n a_{ n } x^{ n+1 } + \sum_{ n = 1 }^{ \infty }\,-8n a_{ n } x^{ n-1 }\\ (x^2 - 1)y'' &= \sum_{ n = 2 }^{ \infty }\,n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,-n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ j = n-1 $ and $ p = n+1 $ to make all terms of the form $ x^{\text{index}} $ rather than the $ x^{\text{index}-1} $ or $ x^{\text{index}+2} $ or whatever. $$ \begin{align} 8y&= \sum_{ n = 0 }^{ \infty }\,8 a_{ n } x^{ n }\\ (-x^2 - 8)y'&= \sum_{ p = 2 }^{ \infty }\,-(p-1) a_{ p-1 } x^{ p } + \sum_{ j = 0 }^{ \infty }\,-8(j+1) a_{ j+1 } x^{ j }\\ (x^2 - 1)y''&= \sum_{ n = 2 }^{ \infty }\,n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,-(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (x^2 - 1)y'' + (-x^2 - 8)y' + 8y &= \sum_{ m = 2 }^{ \infty }\,m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,-(m-1) a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-8(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,8 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(-2a_2 - 8a_1 + 8a_0) + (-6a_{3} - 16a_{2} + 8a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-(m+2)(m+1)a_{m+2} + m(m-1)a_m - 8(m+1)a_{m+1} - (m-1)a_{m-1} + 8a_m)x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{8a_1 - 8a_0}{-2} $$ Equating the linear term to 0 we get $$ a_3 = \frac{16a_2 - 8a_1}{-6} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{8(m+1)a_{m+1} - (m(m-1) + 8)a_m + (m-1)a_{m-1}}{-(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 2 $ and $ a_1 = y'(0) = 5.$ We then plug these values into the formulas found in step 5 to compute the coefficients of the solution.

From the equation for the constant term we get $$ a_2 = \frac{8(5) - 8(2)}{-2} = -12 $$ From the equation for the linear term we get $$ a_3 = \frac{16(-12) - 8(5)}{-6} = 116/3 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{8(2+1)(116/3) - (2(2-1) + 8)(-12) + (2-1)(5)}{-(4)(3)} = -351/4 $$

So our solution is $$ y(x) = 2 + 5x - 12x^2 + (116/3)x^3 - (351/4)x^4 + \cdots $$

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