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

Equating the constant term to 0 we get $$ a_2 = \frac{4a_1 + 2a_0}{-8} $$ Equating the linear term to 0 we get $$ a_3 = \frac{10a_2 + 2a_1}{-24} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-(-(m+1)m - 4(m+1))a_{m+1} + 2a_m - 5(m-1)a_{m-1}}{-4(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 3 $ and $ a_1 = y'(0) = 3.$ 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 $$ a_2 = \frac{4(3) + 2(3)}{-8} = -9/4 $$ From the equation for the linear term we get $$ a_3 = \frac{10(-9/4) + 2(3)}{-24} = 11/16 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-(-(2+1)2 - 4(2+1))(11/16) + 2(-9/4) - 5(2-1)(3)}{-4(4)(3)} = 19/128 $$

So our solution is $$ y(x) = 3 + 3x - (9/4)x^2 + (11/16)x^3 + (19/128)x^4 + \cdots $$

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