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 $$ \begin{align} (6x - 1)y &= \sum_{ n = 0 }^{ \infty }\,6 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n }\\ 7y' &= \sum_{ n = 1 }^{ \infty }\,7n a_{ n } x^{ n-1 }\\ (-4x^2 - 1)y'' &= \sum_{ n = 2 }^{ \infty }\,-4n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,-n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ j = n-1 $ and $ 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. $$ \begin{align} (6x - 1)y&= \sum_{ p = 1 }^{ \infty }\,6 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n }\\ 7y'&= \sum_{ j = 0 }^{ \infty }\,7(j+1) a_{ j+1 } x^{ j }\\ (-4x^2 - 1)y''&= \sum_{ n = 2 }^{ \infty }\,-4n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,-(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (-4x^2 - 1)y'' + 7y' + (6x - 1)y &= \sum_{ m = 2 }^{ \infty }\,-4m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,7(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,6 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,- a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(-2a_2 + 7a_1 - a_0) + (-6a_{3} + 14a_{2} - a_{1} + 6a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-(m+2)(m+1)a_{m+2} - 4m(m-1)a_m + 7(m+1)a_{m+1} - a_m + 6a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-7a_1 + a_0}{-2} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-14a_2 + a_1 - 6a_0}{-6} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-7(m+1)a_{m+1} - (-4m(m-1) - 1)a_m - 6a_{m-1}}{-(m+2)(m+1)} $$ 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 $$ a_2 = \frac{-7(-3) + (-6)}{-2} = -15/2 $$ From the equation for the linear term we get $$ a_3 = \frac{-14(-15/2) + (-3) - 6(-6)}{-6} = -23 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-7(2+1)(-23) - (-42(2-1) - 1)(-15/2) - 6(-3)}{-(4)(3)} = -289/8 $$

So our solution is $$ y(x) = -6 - 3x - (15/2)x^2 - 23x^3 - (289/8)x^4 + \cdots $$

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