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

Equating the constant term to 0 we get $$ a_2 = \frac{-5a_0}{-4} $$ Equating the linear term to 0 we get $$ a_3 = \frac{-4a_2 - 11a_1}{-12} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-2(m+1)ma_{m+1} - (6m + 5)a_m}{-2(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 2 $ and $ a_1 = y'(0) = 0.$ 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 $$ a_2 = \frac{-5(2)}{-4} = 5/2 $$ From the equation for the linear term we get $$ a_3 = \frac{-4(5/2) - 11(0)}{-12} = 5/6 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-2(2+1)2(5/6) - (62 + 5)(5/2)}{-2(4)(3)} = 35/16 $$

So our solution is $$ y(x) = 2 + (5/2)x^2 + (5/6)x^3 + (35/16)x^4 + \cdots $$

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