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 $$ \begin{align} (-5x^2 + 2)y &= \sum_{ n = 0 }^{ \infty }\,-5 a_{ n } x^{ n+2 } + \sum_{ n = 0 }^{ \infty }\,2 a_{ n } x^{ n }\\ (4x + 5)y' &= \sum_{ n = 1 }^{ \infty }\,4n a_{ n } x^{ n } + \sum_{ n = 1 }^{ \infty }\,5n a_{ n } x^{ n-1 }\\ 4y'' &= \sum_{ n = 2 }^{ \infty }\,4n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} (-5x^2 + 2)y&= \sum_{ q = 2 }^{ \infty }\,-5 a_{ q-2 } x^{ q } + \sum_{ n = 0 }^{ \infty }\,2 a_{ n } x^{ n }\\ (4x + 5)y'&= \sum_{ n = 1 }^{ \infty }\,4n a_{ n } x^{ n } + \sum_{ j = 0 }^{ \infty }\,5(j+1) a_{ j+1 } x^{ j }\\ 4y''&= \sum_{ k = 0 }^{ \infty }\,4(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} 4y'' + (4x + 5)y' + (-5x^2 + 2)y &= \sum_{ m = 0 }^{ \infty }\,4(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,4m a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,5(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,-5 a_{ m-2 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,2 a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(8a_2 + 5a_1 + 2a_0) + (24a_{3} + 10a_{2} + 4a_{1} + 2a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(4(m+2)(m+1)a_{m+2} + 5(m+1)a_{m+1} + 4ma_m + 2a_m - 5a_{m-2})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{-5a_1 - 2a_0}{8} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-10a_2 - 6a_1}{24} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-5(m+1)a_{m+1} - (4m + 2)a_m + 5a_{m-2}}{4(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 2 $ and $ a_1 = y'(0) = 3.$ 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 $$ a_2 = \frac{-5(3) - 2(2)}{8} = -19/8 $$ From the equation for the linear term we get $$ a_3 = \frac{-10(-19/8) - 6(3)}{24} = 23/96 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-5(2+1)(23/96) - (42 + 2)(-19/8) + 5(2)}{4(4)(3)} = 965/1536 $$

So our solution is $$ y(x) = 2 + 3x - (19/8)x^2 + (23/96)x^3 + (965/1536)x^4 + \cdots $$

You may reload this page to generate additional examples.