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

Equating the constant term to 0 we get $$ a_2 = \frac{-6a_1 - a_0}{-4} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-12a_2 - a_1}{-12} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-6(m+1)a_{m+1} - (5m(m-1) + 1)a_m - 6(m-1)a_{m-1}}{-2(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -2 $ and $ a_1 = y'(0) = -6.$ 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 $$ a_2 = \frac{-6(-6) - (-2)}{-4} = -19/2 $$ From the equation for the linear term we get $$ a_3 = \frac{-12(-19/2) - (-6)}{-12} = -10 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-6(2+1)(-10) - (52(2-1) + 1)(-19/2) - 6(2-1)(-6)}{-2(4)(3)} = -641/48 $$

So our solution is $$ y(x) = -2 - 6x - (19/2)x^2 - 10x^3 - (641/48)x^4 + \cdots $$

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