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 $$ \begin{align} (-3x - 7)y &= \sum_{ n = 0 }^{ \infty }\,-3 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ (2x^2 + 4)y' &= \sum_{ n = 1 }^{ \infty }\,2n a_{ n } x^{ n+1 } + \sum_{ n = 1 }^{ \infty }\,4n a_{ n } x^{ n-1 }\\ 6y'' &= \sum_{ n = 2 }^{ \infty }\,6n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (-3x - 7)y&= \sum_{ p = 1 }^{ \infty }\,-3 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ (2x^2 + 4)y'&= \sum_{ p = 2 }^{ \infty }\,2(p-1) a_{ p-1 } x^{ p } + \sum_{ j = 0 }^{ \infty }\,4(j+1) a_{ j+1 } x^{ j }\\ 6y''&= \sum_{ k = 0 }^{ \infty }\,6(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} 6y'' + (2x^2 + 4)y' + (-3x - 7)y &= \sum_{ m = 0 }^{ \infty }\,6(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,2(m-1) a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,4(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,-3 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(12a_2 + 4a_1 - 7a_0) + (36a_{3} + 8a_{2} - 7a_{1} - 3a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(6(m+2)(m+1)a_{m+2} + 4(m+1)a_{m+1} + 2(m-1)a_{m-1} - 7a_m - 3a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-4a_1 + 7a_0}{12} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-8a_2 + 7a_1 + 3a_0}{36} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-4(m+1)a_{m+1} + 7a_m - (2(m-1)a_m - 3a_m)a_{m-1}}{6(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -3 $ and $ a_1 = y'(0) = -7.$ 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 $$ a_2 = \frac{-4(-7) + 7(-3)}{12} = 7/12 $$ From the equation for the linear term we get $$ a_3 = \frac{-8(7/12) + 7(-7) + 3(-3)}{36} = -47/27 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-4(2+1)(-47/27) + 7(7/12) - (2(2-1)(7/12) - 3(7/12))(-7)}{6(4)(3)} = 647/2592 $$

So our solution is $$ y(x) = -3 - 7x + (7/12)x^2 - (47/27)x^3 + (647/2592)x^4 + \cdots $$

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