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 $$ \begin{align} 3y &= \sum_{ n = 0 }^{ \infty }\,3 a_{ n } x^{ n }\\ (-3x + 8)y' &= \sum_{ n = 1 }^{ \infty }\,-3n a_{ n } x^{ n } + \sum_{ n = 1 }^{ \infty }\,8n a_{ n } x^{ n-1 }\\ (-4x^2 + 6)y'' &= \sum_{ n = 2 }^{ \infty }\,-4n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,6n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} 3y&= \sum_{ n = 0 }^{ \infty }\,3 a_{ n } x^{ n }\\ (-3x + 8)y'&= \sum_{ n = 1 }^{ \infty }\,-3n a_{ n } x^{ n } + \sum_{ j = 0 }^{ \infty }\,8(j+1) a_{ j+1 } x^{ j }\\ (-4x^2 + 6)y''&= \sum_{ n = 2 }^{ \infty }\,-4n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,6(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (-4x^2 + 6)y'' + (-3x + 8)y' + 3y &= \sum_{ m = 2 }^{ \infty }\,-4m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,6(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,-3m a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,8(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,3 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(12a_2 + 8a_1 + 3a_0) + (36a_{3} + 16a_{2} - 3a_{1} + 3a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(6(m+2)(m+1)a_{m+2} - 4m(m-1)a_m + 8(m+1)a_{m+1} - 3ma_m + 3a_m)x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-8a_1 - 3a_0}{12} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-16a_2}{36} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-8(m+1)a_{m+1} - (-4m(m-1) - 3m + 3)a_m}{6(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -4 $ 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 $$ a_2 = \frac{-8(8) - 3(-4)}{12} = -13/3 $$ From the equation for the linear term we get $$ a_3 = \frac{-16(-13/3)}{36} = 52/27 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-8(2+1)(52/27) - (-42(2-1) - 32 + 3)(-13/3)}{6(4)(3)} = -845/648 $$

So our solution is $$ y(x) = -4 + 8x - (13/3)x^2 + (52/27)x^3 - (845/648)x^4 + \cdots $$

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