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 $$ \begin{align} (5x + 4)y &= \sum_{ n = 0 }^{ \infty }\,5 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,4 a_{ n } x^{ n }\\ 8y' &= \sum_{ n = 1 }^{ \infty }\,8n a_{ n } x^{ n-1 }\\ (5x - 1)y'' &= \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n-1 } + \sum_{ n = 2 }^{ \infty }\,-n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (5x + 4)y&= \sum_{ p = 1 }^{ \infty }\,5 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,4 a_{ n } x^{ n }\\ 8y'&= \sum_{ j = 0 }^{ \infty }\,8(j+1) a_{ j+1 } x^{ j }\\ (5x - 1)y''&= \sum_{ j = 1 }^{ \infty }\,5(j+1)j a_{ j+1 } x^{ j } + \sum_{ k = 0 }^{ \infty }\,-(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (5x - 1)y'' + 8y' + (5x + 4)y &= \sum_{ m = 1 }^{ \infty }\,5(m+1)m a_{ m+1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,8(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,5 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,4 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. (While the general term starts at m=1, it doesn't hurt to separate out the x term as we do below). $$ \begin{align} &(-2a_2 + 8a_1 + 4a_0) + (-6a_{3} + 10a_{2} + 16a_{2} + 4a_{1} + 5a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-(m+2)(m+1)a_{m+2} + 5(m+1)ma_{m+1} + 8(m+1)a_{m+1} + 4a_m + 5a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-8a_1 - 4a_0}{-2} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-26a_2 - 4a_1 - 5a_0}{-6} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-(5(m+1)m + 8(m+1))a_{m+1} - 4a_m - 5a_{m-1}}{-(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 6 $ 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 $$ a_2 = \frac{-8(4) - 4(6)}{-2} = 28 $$ From the equation for the linear term we get $$ a_3 = \frac{-26(28) - 4(4) - 5(6)}{-6} = 129 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-(5(2+1)2 + 8(2+1))(129) - 4(28) - 5(4)}{-(4)(3)} = 1183/2 $$

So our solution is $$ y(x) = 6 + 4x + 28x^2 + 129x^3 + (1183/2)x^4 + \cdots $$

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