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Series Solutions

Additional Examples

Find the series solution (up to the $ x^4 $ term) for the equation $$ (-3x + 5)y'' + (-3x^2 - 1)y = 0,\qquad y(0) = 6, y'(0) = -2 $$

STEP 1: Plug in $\displaystyle y(x) = \sum_{n=0}^{\infty}a_n x^n $ and compute all the different terms in the equation $$ \begin{align} (-3x^2 - 1)y &= \sum_{ n = 0 }^{ \infty }\,-3 a_{ n } x^{ n+2 } + \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n }\\ y' &= \\ (-3x + 5)y'' &= \sum_{ n = 2 }^{ \infty }\,-3n(n-1) a_{ n } x^{ n-1 } + \sum_{ n = 2 }^{ \infty }\,5n(n-1) a_{ n } x^{ n-2 } \end{align} $$ STEP 2: Make the substitutions $ k = n-2 $, $ 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. $$ \begin{align} (-3x^2 - 1)y&= \sum_{ q = 2 }^{ \infty }\,-3 a_{ q-2 } x^{ q } + \sum_{ n = 0 }^{ \infty }\,- a_{ n } x^{ n }\\ y'&= \\ (-3x + 5)y''&= \sum_{ j = 1 }^{ \infty }\,-3(j+1)j a_{ j+1 } x^{ j } + \sum_{ k = 0 }^{ \infty }\,5(k+2)(k+1) a_{ k+2 } x^{ k }\end{align} $$ STEP 3: Change all the indices to the same letter (I use $ m $) and plug into the equation. $$ \begin{align} (-3x + 5)y'' + (-3x^2 - 1)y &= \sum_{ m = 1 }^{ \infty }\,-3(m+1)m a_{ m+1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,5(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \\ &+ \sum_{ m = 2 }^{ \infty }\,-3 a_{ m-2 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,- a_{ m } x^{ m } \end{align} $$ STEP 4: Collect like terms. $$ \begin{align} &(10a_2 - a_0) + (30a_{3} - 6a_{2} - a_{1})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(5(m+2)(m+1)a_{m+2} - 3(m+1)ma_{m+1} - a_m - 3a_{m-2})x^m = 0 \end{align} $$ STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$ a_2 = \frac{a_0}{10} $$ Equating the linear term to 0 we get $$ a_3 = \frac{6a_2 + a_1}{30} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{3(m+1)ma_{m+1} + a_m + 3a_{m-2}}{5(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 6 $ and $ a_1 = y'(0) = -2.$ 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 $$ a_2 = \frac{(6)}{10} = 3/5 $$ From the equation for the linear term we get $$ a_3 = \frac{6(3/5) + (-2)}{30} = 4/75 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{3(2+1)2(4/75) + (3/5) + 3(6)}{5(4)(3)} = 163/500 $$

So our solution is $$ y(x) = 6 - 2x + (3/5)x^2 + (4/75)x^3 + (163/500)x^4 + \cdots $$

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