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

Equating the constant term to 0 we get $$ a_2 = \frac{-8a_1 + 5a_0}{8} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-16a_2}{24} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-8(m+1)a_{m+1} - (3m(m-1) + 5m - 5)a_m}{4(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -7 $ and $ a_1 = y'(0) = 2.$ 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 $$ a_2 = \frac{-8(2) + 5(-7)}{8} = -51/8 $$ From the equation for the linear term we get $$ a_3 = \frac{-16(-51/8)}{24} = 17/4 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-8(2+1)(17/4) - (32(2-1) + 52 - 5)(-51/8)}{4(4)(3)} = -85/128 $$

So our solution is $$ y(x) = -7 + 2x - (51/8)x^2 + (17/4)x^3 - (85/128)x^4 + \cdots $$

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