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 $$ \begin{align} (-x^2 + 1)y &= \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n+2 } + \sum_{ n = 0 }^{ \infty }\, a_{ n } x^{ n }\\ 4y' &= \sum_{ n = 1 }^{ \infty }\,4n a_{ n } x^{ n-1 }\\ (6x^2 - 2)y'' &= \sum_{ n = 2 }^{ \infty }\,6n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,-2n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ j = n-1 $ and $ 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. $$ \begin{align} (-x^2 + 1)y&= \sum_{ q = 2 }^{ \infty }\,- a_{ q-2 } x^{ q } + \sum_{ n = 0 }^{ \infty }\, a_{ n } x^{ n }\\ 4y'&= \sum_{ j = 0 }^{ \infty }\,4(j+1) a_{ j+1 } x^{ j }\\ (6x^2 - 2)y''&= \sum_{ n = 2 }^{ \infty }\,6n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,-2(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (6x^2 - 2)y'' + 4y' + (-x^2 + 1)y &= \sum_{ m = 2 }^{ \infty }\,6m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-2(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,4(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,- a_{ m-2 } x^{ m } + \sum_{ m = 0 }^{ \infty }\, a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(-4a_2 + 4a_1 + a_0) + (-12a_{3} + 8a_{2} + a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-2(m+2)(m+1)a_{m+2} + 6m(m-1)a_m + 4(m+1)a_{m+1} + a_m - a_{m-2})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-4a_1 - a_0}{-4} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-8a_2 - a_1}{-12} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-4(m+1)a_{m+1} - (6m(m-1) + 1)a_m + a_{m-2}}{-2(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -6 $ and $ a_1 = y'(0) = 3.$ 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 $$ a_2 = \frac{-4(3) - (-6)}{-4} = 3/2 $$ From the equation for the linear term we get $$ a_3 = \frac{-8(3/2) - (3)}{-12} = 5/4 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-4(2+1)(5/4) - (62(2-1) + 1)(3/2) + (-6)}{-2(4)(3)} = 27/16 $$

So our solution is $$ y(x) = -6 + 3x + (3/2)x^2 + (5/4)x^3 + (27/16)x^4 + \cdots $$

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