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 $$ \begin{align} (-x^2 + 7)y &= \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n+2 } + \sum_{ n = 0 }^{ \infty }\,7 a_{ n } x^{ n }\\ -1y' &= \sum_{ n = 1 }^{ \infty }\,-n a_{ n } x^{ n-1 }\\ (3x^2 + 3)y'' &= \sum_{ n = 2 }^{ \infty }\,3n(n-1) a_{ n } x^{ n } + \sum_{ n = 2 }^{ \infty }\,3n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ j = n-1 $ and $ q = n+2 $ to make all terms of the form $ x^{\text{index}} $ rather than the $ x^{\text{index}-1} $ or $ x^{\text{index}+2} $ or whatever. $$ \begin{align} (-x^2 + 7)y&= \sum_{ q = 2 }^{ \infty }\,- a_{ q-2 } x^{ q } + \sum_{ n = 0 }^{ \infty }\,7 a_{ n } x^{ n }\\ -1y'&= \sum_{ j = 0 }^{ \infty }\,-(j+1) a_{ j+1 } x^{ j }\\ (3x^2 + 3)y''&= \sum_{ n = 2 }^{ \infty }\,3n(n-1) a_{ n } x^{ n } + \sum_{ k = 0 }^{ \infty }\,3(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (3x^2 + 3)y'' - 1y' + (-x^2 + 7)y &= \sum_{ m = 2 }^{ \infty }\,3m(m-1) a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,3(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,-(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,- a_{ m-2 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,7 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(6a_2 - a_1 + 7a_0) + (18a_{3} - 2a_{2} + 7a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(3(m+2)(m+1)a_{m+2} + 3m(m-1)a_m - (m+1)a_{m+1} + 7a_m - a_{m-2})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{a_1 - 7a_0}{6} $$ Equating the linear term to 0 we get $$ a_3 = \frac{2a_2 - 7a_1}{18} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{(m+1)a_{m+1} - (3m(m-1) + 7)a_m + a_{m-2}}{3(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 1 $ and $ a_1 = y'(0) = -6.$ 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 $$ a_2 = \frac{(-6) - 7(1)}{6} = -13/6 $$ From the equation for the linear term we get $$ a_3 = \frac{2(-13/6) - 7(-6)}{18} = 113/54 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{(2+1)(113/54) - (32(2-1) + 7)(-13/6) + (1)}{3(4)(3)} = 319/324 $$

So our solution is $$ y(x) = 1 - 6x - (13/6)x^2 + (113/54)x^3 + (319/324)x^4 + \cdots $$

You may reload this page to generate additional examples.