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 $$ \begin{align} 4y &= \sum_{ n = 0 }^{ \infty }\,4 a_{ n } x^{ n }\\ (-x + 7)y' &= \sum_{ n = 1 }^{ \infty }\,-n a_{ n } x^{ n } + \sum_{ n = 1 }^{ \infty }\,7n a_{ n } x^{ n-1 }\\ (4x^2 - 6)y'' &= \sum_{ n = 2 }^{ \infty }\,4n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,-6n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} 4y&= \sum_{ n = 0 }^{ \infty }\,4 a_{ n } x^{ n }\\ (-x + 7)y'&= \sum_{ n = 1 }^{ \infty }\,-n a_{ n } x^{ n } + \sum_{ j = 0 }^{ \infty }\,7(j+1) a_{ j+1 } x^{ j }\\ (4x^2 - 6)y''&= \sum_{ n = 2 }^{ \infty }\,4n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,-6(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (4x^2 - 6)y'' + (-x + 7)y' + 4y &= \sum_{ m = 2 }^{ \infty }\,4m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-6(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,-m a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,7(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,4 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(-12a_2 + 7a_1 + 4a_0) + (-36a_{3} + 14a_{2} - a_{1} + 4a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-6(m+2)(m+1)a_{m+2} + 4m(m-1)a_m + 7(m+1)a_{m+1} - ma_m + 4a_m)x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-7a_1 - 4a_0}{-12} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-14a_2 - 3a_1}{-36} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-7(m+1)a_{m+1} - (4m(m-1) - m + 4)a_m}{-6(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 7 $ 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 $$ a_2 = \frac{-7(-2) - 4(7)}{-12} = 7/6 $$ From the equation for the linear term we get $$ a_3 = \frac{-14(7/6) - 3(-2)}{-36} = 31/108 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-7(2+1)(31/108) - (42(2-1) - 2 + 4)(7/6)}{-6(4)(3)} = 637/2592 $$

So our solution is $$ y(x) = 7 - 2x + (7/6)x^2 + (31/108)x^3 + (637/2592)x^4 + \cdots $$

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