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 $$ \begin{align} 8y &= \sum_{ n = 0 }^{ \infty }\,8 a_{ n } x^{ n }\\ 6xy' &= \sum_{ n = 1 }^{ \infty }\,6n a_{ n } x^{ n }\\ (5x^2 + 5)y'' &= \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitution $ k = 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. $$ \begin{align} 8y&= \sum_{ n = 0 }^{ \infty }\,8 a_{ n } x^{ n }\\ 6xy'&= \sum_{ n = 1 }^{ \infty }\,6n a_{ n } x^{ n }\\ (5x^2 + 5)y''&= \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,5(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (5x^2 + 5)y'' + 6xy' + 8y &= \sum_{ m = 2 }^{ \infty }\,5m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,5(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,6m a_{ m } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,8 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(10a_2 + 8a_0) + (30a_{3} + 6a_{1} + 8a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(5(m+2)(m+1)a_{m+2} + 5m(m-1)a_m + 6ma_m + 8a_m)x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-8a_0}{10} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-14a_1}{30} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-(5m(m-1) + 6m + 8)a_m}{5(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -2 $ and $ a_1 = y'(0) = -4.$ 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 $$ a_2 = \frac{-8(-2)}{10} = 8/5 $$ From the equation for the linear term we get $$ a_3 = \frac{-14(-4)}{30} = 28/15 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-(52(2-1) + 62 + 8)(8/5)}{5(4)(3)} = -4/5 $$

So our solution is $$ y(x) = -2 - 4x + (8/5)x^2 + (28/15)x^3 - (4/5)x^4 + \cdots $$

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