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 $$ \begin{align} (-x - 7)y &= \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ 8y' &= \sum_{ n = 1 }^{ \infty }\,8n a_{ n } x^{ n-1 }\\ (5x^2 - 4)y'' &= \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,-4n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (-x - 7)y&= \sum_{ p = 1 }^{ \infty }\,- a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ 8y'&= \sum_{ j = 0 }^{ \infty }\,8(j+1) a_{ j+1 } x^{ j }\\ (5x^2 - 4)y''&= \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,-4(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (5x^2 - 4)y'' + 8y' + (-x - 7)y &= \sum_{ m = 2 }^{ \infty }\,5m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-4(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,8(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,- a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(-8a_2 + 8a_1 - 7a_0) + (-24a_{3} + 16a_{2} - 7a_{1} - a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-4(m+2)(m+1)a_{m+2} + 5m(m-1)a_m + 8(m+1)a_{m+1} - 7a_m - a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-8a_1 + 7a_0}{-8} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-16a_2 + 7a_1 + a_0}{-24} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-8(m+1)a_{m+1} - (5m(m-1) - 7)a_m + a_{m-1}}{-4(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -1 $ and $ a_1 = y'(0) = 1.$ 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 $$ a_2 = \frac{-8(1) + 7(-1)}{-8} = 15/8 $$ From the equation for the linear term we get $$ a_3 = \frac{-16(15/8) + 7(1) + (-1)}{-24} = 1 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-8(2+1)(1) - (52(2-1) - 7)(15/8) + (1)}{-4(4)(3)} = 229/384 $$

So our solution is $$ y(x) = -1 + x + (15/8)x^2 + x^3 + (229/384)x^4 + \cdots $$

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