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 $$ \begin{align} (-4x + 8)y &= \sum_{ n = 0 }^{ \infty }\,-4 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,8 a_{ n } x^{ n }\\ (2x + 7)y' &= \sum_{ n = 1 }^{ \infty }\,2n a_{ n } x^{ n } + \sum_{ n = 1 }^{ \infty }\,7n a_{ n } x^{ n-1 }\\ -3y'' &= \sum_{ n = 2 }^{ \infty }\,-3n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (-4x + 8)y&= \sum_{ p = 1 }^{ \infty }\,-4 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,8 a_{ n } x^{ n }\\ (2x + 7)y'&= \sum_{ n = 1 }^{ \infty }\,2n a_{ n } x^{ n } + \sum_{ j = 0 }^{ \infty }\,7(j+1) a_{ j+1 } x^{ j }\\ -3y''&= \sum_{ k = 0 }^{ \infty }\,-3(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} -3y'' + (2x + 7)y' + (-4x + 8)y &= \sum_{ m = 0 }^{ \infty }\,-3(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,2m a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,7(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,-4 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,8 a_{ m } x^{ m } \end{align} $$ 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). $$ \begin{align} &(-6a_2 + 7a_1 + 8a_0) + (-18a_{3} + 14a_{2} + 2a_{1} + 8a_{1} - 4a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-3(m+2)(m+1)a_{m+2} + 7(m+1)a_{m+1} + 2ma_m + 8a_m - 4a_{m-1})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-7a_1 - 8a_0}{-6} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-14a_2 - 10a_1 + 4a_0}{-18} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-7(m+1)a_{m+1} - (2m + 8)a_m + 4a_{m-1}}{-3(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 5 $ 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 $$ a_2 = \frac{-7(-2) - 8(5)}{-6} = 13/3 $$ From the equation for the linear term we get $$ a_3 = \frac{-14(13/3) - 10(-2) + 4(5)}{-18} = 31/27 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-7(2+1)(31/27) - (22 + 8)(13/3) + 4(-2)}{-3(4)(3)} = 757/324 $$

So our solution is $$ y(x) = 5 - 2x + (13/3)x^2 + (31/27)x^3 + (757/324)x^4 + \cdots $$

You may reload this page to generate additional examples.