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 $$ \begin{align} (4x + 7)y &= \sum_{ n = 0 }^{ \infty }\,4 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,7 a_{ n } x^{ n }\\ 5x^2y' &= \sum_{ n = 1 }^{ \infty }\,5n a_{ n } x^{ n+1 }\\ 2y'' &= \sum_{ n = 2 }^{ \infty }\,2n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $ 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. $$ \begin{align} (4x + 7)y&= \sum_{ p = 1 }^{ \infty }\,4 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,7 a_{ n } x^{ n }\\ 5x^2y'&= \sum_{ p = 2 }^{ \infty }\,5(p-1) a_{ p-1 } x^{ p }\\ 2y''&= \sum_{ k = 0 }^{ \infty }\,2(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} 2y'' + 5x^2y' + (4x + 7)y &= \sum_{ m = 0 }^{ \infty }\,2(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,5(m-1) a_{ m-1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,4 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(4a_2 + 7a_0) + (12a_{3} + 7a_{1} + 4a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(2(m+2)(m+1)a_{m+2} + 5(m-1)a_{m-1} + 7a_m + 4a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-7a_0}{4} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-7a_1 - 4a_0}{12} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-7a_m - (5(m-1)a_m + 4a_m)a_{m-1}}{2(m+2)(m+1)} $$ 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 $$ a_2 = \frac{-7(1)}{4} = -7/4 $$ From the equation for the linear term we get $$ a_3 = \frac{-7(-2) - 4(1)}{12} = 5/6 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-7(-7/4) - (5(2-1)(-7/4) + 4(-7/4))(-2)}{2(4)(3)} = 121/96 $$

So our solution is $$ y(x) = 1 - 2x - (7/4)x^2 + (5/6)x^3 + (121/96)x^4 + \cdots $$

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