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 $$ \begin{align} (-5x - 7)y &= \sum_{ n = 0 }^{ \infty }\,-5 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ -8y' &= \sum_{ n = 1 }^{ \infty }\,-8n a_{ n } x^{ n-1 }\\ (4x - 3)y'' &= \sum_{ n = 2 }^{ \infty }\,4n(n-1) a_{ n } x^{ n-1 } + \sum_{ n = 2 }^{ \infty }\,-3n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (-5x - 7)y&= \sum_{ p = 1 }^{ \infty }\,-5 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ -8y'&= \sum_{ j = 0 }^{ \infty }\,-8(j+1) a_{ j+1 } x^{ j }\\ (4x - 3)y''&= \sum_{ j = 1 }^{ \infty }\,4(j+1)j a_{ j+1 } x^{ j } + \sum_{ k = 0 }^{ \infty }\,-3(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (4x - 3)y'' - 8y' + (-5x - 7)y &= \sum_{ m = 1 }^{ \infty }\,4(m+1)m a_{ m+1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-3(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,-8(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,-5 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. (While the general term starts at m=1, it doesn't hurt to separate out the x term as we do below). $$ \begin{align} &(-6a_2 - 8a_1 - 7a_0) + (-18a_{3} + 8a_{2} - 16a_{2} - 7a_{1} - 5a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-3(m+2)(m+1)a_{m+2} + 4(m+1)ma_{m+1} - 8(m+1)a_{m+1} - 7a_m - 5a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{8a_1 + 7a_0}{-6} $$ Equating the linear term to 0 we get $$ a_3 = \frac{8a_2 + 7a_1 + 5a_0}{-18} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-(4(m+1)m - 8(m+1))a_{m+1} + 7a_m + 5a_{m-1}}{-3(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 1 $ 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 $$ a_2 = \frac{8(2) + 7(1)}{-6} = -23/6 $$ From the equation for the linear term we get $$ a_3 = \frac{8(-23/6) + 7(2) + 5(1)}{-18} = 35/54 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-(4(2+1)2 - 8(2+1))(35/54) + 7(-23/6) + 5(2)}{-3(4)(3)} = 101/216 $$

So our solution is $$ y(x) = 1 + 2x - (23/6)x^2 + (35/54)x^3 + (101/216)x^4 + \cdots $$

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