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

Equating the constant term to 0 we get $$ a_2 = \frac{4a_1 + 3a_0}{2} $$ Equating the linear term to 0 we get $$ a_3 = \frac{8a_2 + a_1}{6} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{4(m+1)a_{m+1} - (2m - 3)a_m + 4a_{m-2}}{(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 6 $ and $ a_1 = y'(0) = 5.$ 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 $$ a_2 = \frac{4(5) + 3(6)}{2} = 19 $$ From the equation for the linear term we get $$ a_3 = \frac{8(19) + (5)}{6} = 157/6 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{4(2+1)(157/6) - (22 - 3)(19) + 4(6)}{(4)(3)} = 319/12 $$

So our solution is $$ y(x) = 6 + 5x + 19x^2 + (157/6)x^3 + (319/12)x^4 + \cdots $$

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