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

Equating the constant term to 0 we get $$ a_2 = \frac{2a_1 - 4a_0}{12} $$ Equating the linear term to 0 we get $$ a_3 = \frac{4a_2 + a_1}{36} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{2(m+1)a_{m+1} - (-3m(m-1) - 5m + 4)a_m}{6(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 2 $ and $ a_1 = y'(0) = -8.$ 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 $$ a_2 = \frac{2(-8) - 4(2)}{12} = -2 $$ From the equation for the linear term we get $$ a_3 = \frac{4(-2) + (-8)}{36} = -4/9 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{2(2+1)(-4/9) - (-32(2-1) - 52 + 4)(-2)}{6(4)(3)} = -10/27 $$

So our solution is $$ y(x) = 2 - 8x - 2x^2 - (4/9)x^3 - (10/27)x^4 + \cdots $$

You may reload this page to generate additional examples.