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 $$ \begin{align} (-3x^2 - 7)y &= \sum_{ n = 0 }^{ \infty }\,-3 a_{ n } x^{ n+2 } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ 7y' &= \sum_{ n = 1 }^{ \infty }\,7n a_{ n } x^{ n-1 }\\ (x + 4)y'' &= \sum_{ n = 2 }^{ \infty }\,n(n-1) a_{ n } x^{ n-1 } + \sum_{ n = 2 }^{ \infty }\,4n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ j = n-1 $ and $ q = 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. $$ \begin{align} (-3x^2 - 7)y&= \sum_{ q = 2 }^{ \infty }\,-3 a_{ q-2 } x^{ q } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ 7y'&= \sum_{ j = 0 }^{ \infty }\,7(j+1) a_{ j+1 } x^{ j }\\ (x + 4)y''&= \sum_{ j = 1 }^{ \infty }\,(j+1)j a_{ j+1 } x^{ j } + \sum_{ k = 0 }^{ \infty }\,4(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (x + 4)y'' + 7y' + (-3x^2 - 7)y &= \sum_{ m = 1 }^{ \infty }\,(m+1)m a_{ m+1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,4(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,7(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,-3 a_{ m-2 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(8a_2 + 7a_1 - 7a_0) + (24a_{3} + 2a_{2} + 14a_{2} - 7a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(4(m+2)(m+1)a_{m+2} + (m+1)ma_{m+1} + 7(m+1)a_{m+1} - 7a_m - 3a_{m-2})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-7a_1 + 7a_0}{8} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-16a_2 + 7a_1}{24} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-((m+1)m + 7(m+1))a_{m+1} + 7a_m + 3a_{m-2}}{4(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 6 $ and $ a_1 = y'(0) = 8.$ 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 $$ a_2 = \frac{-7(8) + 7(6)}{8} = -7/4 $$ From the equation for the linear term we get $$ a_3 = \frac{-16(-7/4) + 7(8)}{24} = 7/2 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-((2+1)2 + 7(2+1))(7/2) + 7(-7/4) + 3(6)}{4(4)(3)} = -355/192 $$

So our solution is $$ y(x) = 6 + 8x - (7/4)x^2 + (7/2)x^3 - (355/192)x^4 + \cdots $$

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