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

Equating the constant term to 0 we get $$ a_2 = \frac{-6a_1 + 4a_0}{2} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-12a_2 + 4a_1 - 6a_0}{6} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-6(m+1)a_{m+1} - (4m(m-1) - 4)a_m - 6a_{m-1}}{(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -7 $ and $ a_1 = y'(0) = 0.$ 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 $$ a_2 = \frac{-6(0) + 4(-7)}{2} = -14 $$ From the equation for the linear term we get $$ a_3 = \frac{-12(-14) + 4(0) - 6(-7)}{6} = 35 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-6(2+1)(35) - (42(2-1) - 4)(-14) - 6(0)}{(4)(3)} = -287/6 $$

So our solution is $$ y(x) = -7 - 14x^2 + 35x^3 - (287/6)x^4 + \cdots $$

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