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 $$ \begin{align} (3x - 7)y &= \sum_{ n = 0 }^{ \infty }\,3 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ (2x^2 + 5)y' &= \sum_{ n = 1 }^{ \infty }\,2n a_{ n } x^{ n+1 } + \sum_{ n = 1 }^{ \infty }\,5n a_{ n } x^{ n-1 }\\ -5y'' &= \sum_{ n = 2 }^{ \infty }\,-5n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (3x - 7)y&= \sum_{ p = 1 }^{ \infty }\,3 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,-7 a_{ n } x^{ n }\\ (2x^2 + 5)y'&= \sum_{ p = 2 }^{ \infty }\,2(p-1) a_{ p-1 } x^{ p } + \sum_{ j = 0 }^{ \infty }\,5(j+1) a_{ j+1 } x^{ j }\\ -5y''&= \sum_{ k = 0 }^{ \infty }\,-5(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} -5y'' + (2x^2 + 5)y' + (3x - 7)y &= \sum_{ m = 0 }^{ \infty }\,-5(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,2(m-1) a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,5(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,3 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(-10a_2 + 5a_1 - 7a_0) + (-30a_{3} + 10a_{2} - 7a_{1} + 3a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-5(m+2)(m+1)a_{m+2} + 5(m+1)a_{m+1} + 2(m-1)a_{m-1} - 7a_m + 3a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-5a_1 + 7a_0}{-10} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-10a_2 + 7a_1 - 3a_0}{-30} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-5(m+1)a_{m+1} + 7a_m - (2(m-1)a_m + 3a_m)a_{m-1}}{-5(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -4 $ 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 $$ a_2 = \frac{-5(-7) + 7(-4)}{-10} = -7/10 $$ From the equation for the linear term we get $$ a_3 = \frac{-10(-7/10) + 7(-7) - 3(-4)}{-30} = 1 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-5(2+1)(1) + 7(-7/10) - (2(2-1)(-7/10) + 3(-7/10))(-7)}{-5(4)(3)} = -151/600 $$

So our solution is $$ y(x) = -4 - 7x - (7/10)x^2 + x^3 - (151/600)x^4 + \cdots $$

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