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 $$ \begin{align} 5y &= \sum_{ n = 0 }^{ \infty }\,5 a_{ n } x^{ n }\\ (x - 5)y' &= \sum_{ n = 1 }^{ \infty }\,n a_{ n } x^{ n } + \sum_{ n = 1 }^{ \infty }\,-5n a_{ n } x^{ n-1 }\\ (-3x - 5)y'' &= \sum_{ n = 2 }^{ \infty }\,-3n(n-1) a_{ n } x^{ n-1 } + \sum_{ n = 2 }^{ \infty }\,-5n(n-1) a_{ n } x^{ n-2 } \end{align} $$ 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. $$ \begin{align} 5y&= \sum_{ n = 0 }^{ \infty }\,5 a_{ n } x^{ n }\\ (x - 5)y'&= \sum_{ n = 1 }^{ \infty }\,n a_{ n } x^{ n } + \sum_{ j = 0 }^{ \infty }\,-5(j+1) a_{ j+1 } x^{ j }\\ (-3x - 5)y''&= \sum_{ j = 1 }^{ \infty }\,-3(j+1)j a_{ j+1 } x^{ j } + \sum_{ k = 0 }^{ \infty }\,-5(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (-3x - 5)y'' + (x - 5)y' + 5y &= \sum_{ m = 1 }^{ \infty }\,-3(m+1)m a_{ m+1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-5(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,m a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-5(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 0 }^{ \infty }\,5 a_{ m } x^{ m } \end{align} $$ 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). $$ \begin{align} &(-10a_2 - 5a_1 + 5a_0) + (-30a_{3} - 6a_{2} - 10a_{2} + a_{1} + 5a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-5(m+2)(m+1)a_{m+2} - 3(m+1)ma_{m+1} - 5(m+1)a_{m+1} + ma_m + 5a_m)x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{5a_1 - 5a_0}{-10} $$ Equating the linear term to 0 we get $$ a_3 = \frac{16a_2 - 6a_1}{-30} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{-(-3(m+1)m - 5(m+1))a_{m+1} - (m + 5)a_m}{-5(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = -1 $ 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 $$ a_2 = \frac{5(-3) - 5(-1)}{-10} = 1 $$ From the equation for the linear term we get $$ a_3 = \frac{16(1) - 6(-3)}{-30} = -17/15 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{-(-3(2+1)2 - 5(2+1))(-17/15) - (2 + 5)(1)}{-5(4)(3)} = 37/50 $$

So our solution is $$ y(x) = -1 - 3x + x^2 - (17/15)x^3 + (37/50)x^4 + \cdots $$

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