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 $$ \begin{align} (-6x^2 - 3)y &= \sum_{ n = 0 }^{ \infty }\,-6 a_{ n } x^{ n+2 } + \sum_{ n = 0 }^{ \infty }\,-3 a_{ n } x^{ n }\\ (-5x^2 + 3)y' &= \sum_{ n = 1 }^{ \infty }\,-5n a_{ n } x^{ n+1 } + \sum_{ n = 1 }^{ \infty }\,3n a_{ n } x^{ n-1 }\\ 2y'' &= \sum_{ n = 2 }^{ \infty }\,2n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ j = n-1 $, $ p = n+1 $ and $ q = n+2 $ to make all terms of the form $ x^{\text{index}} $ rather than the $ x^{\text{index}-1} $ or $ x^{\text{index}+2} $ or whatever. $$ \begin{align} (-6x^2 - 3)y&= \sum_{ q = 2 }^{ \infty }\,-6 a_{ q-2 } x^{ q } + \sum_{ n = 0 }^{ \infty }\,-3 a_{ n } x^{ n }\\ (-5x^2 + 3)y'&= \sum_{ p = 2 }^{ \infty }\,-5(p-1) a_{ p-1 } x^{ p } + \sum_{ j = 0 }^{ \infty }\,3(j+1) a_{ j+1 } x^{ j }\\ 2y''&= \sum_{ k = 0 }^{ \infty }\,2(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} 2y'' + (-5x^2 + 3)y' + (-6x^2 - 3)y &= \sum_{ m = 0 }^{ \infty }\,2(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,-5(m-1) a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,3(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,-6 a_{ m-2 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-3 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(4a_2 + 3a_1 - 3a_0) + (12a_{3} + 6a_{2} - 3a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(2(m+2)(m+1)a_{m+2} + 3(m+1)a_{m+1} - 5(m-1)a_{m-1} - 3a_m - 6a_{m-2})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-3a_1 + 3a_0}{4} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-6a_2 + 3a_1}{12} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-3(m+1)a_{m+1} + 3a_m + 5(m-1)a_{m-1} + 6a_{m-2}}{2(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -8 $ and $ a_1 = y'(0) = -8.$ 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 $$ a_2 = \frac{-3(-8) + 3(-8)}{4} = 0 $$ From the equation for the linear term we get $$ a_3 = \frac{-6(0) + 3(-8)}{12} = -2 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-3(2+1)(-2) + 3(0) + 5(2-1)(-8) + 6(-8)}{2(4)(3)} = -35/12 $$

So our solution is $$ y(x) = -8 - 8x - 2x^3 - (35/12)x^4 + \cdots $$

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