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

Equating the constant term to 0 we get $$ a_2 = \frac{-6a_1 + 6a_0}{2} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-12a_2 + 3a_1}{6} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-6(m+1)a_{m+1} - (3m - 6)a_m + 2a_{m-2}}{(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 3 $ and $ a_1 = y'(0) = 1.$ 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 $$ a_2 = \frac{-6(1) + 6(3)}{2} = 6 $$ From the equation for the linear term we get $$ a_3 = \frac{-12(6) + 3(1)}{6} = -23/2 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-6(2+1)(-23/2) - (32 - 6)(6) + 2(3)}{(4)(3)} = 71/4 $$

So our solution is $$ y(x) = 3 + x + 6x^2 - (23/2)x^3 + (71/4)x^4 + \cdots $$

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