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

Equating the constant term to 0 we get $$ a_2 = \frac{-a_1 - 3a_0}{4} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-14a_2 - 3a_1}{12} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-(6(m+1)m + (m+1))a_{m+1} - 3a_m - 2(m-1)a_{m-1}}{2(m+2)(m+1)} $$ 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 $$ a_2 = \frac{-(8) - 3(-8)}{4} = 4 $$ From the equation for the linear term we get $$ a_3 = \frac{-14(4) - 3(8)}{12} = -20/3 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-(6(2+1)2 + (2+1))(-20/3) - 3(4) - 2(2-1)(8)}{2(4)(3)} = 29/3 $$

So our solution is $$ y(x) = -8 + 8x + 4x^2 - (20/3)x^3 + (29/3)x^4 + \cdots $$

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