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### Series Solutions

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

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} (6x - 5)y &= \sum_{ n = 0 }^{ \infty }\,6 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,-5 a_{ n } x^{ n }\\ (5x^2 - 6)y' &= \sum_{ n = 1 }^{ \infty }\,5n a_{ n } x^{ n+1 } + \sum_{ n = 1 }^{ \infty }\,-6n a_{ n } x^{ n-1 }\\ 3y'' &= \sum_{ n = 2 }^{ \infty }\,3n(n-1) a_{ n } x^{ n-2 } \end{align} STEP 2: Make the substitutions $k = n-2$, $j = n-1$ and $p = n+1$ 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} (6x - 5)y&= \sum_{ p = 1 }^{ \infty }\,6 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,-5 a_{ n } x^{ n }\\ (5x^2 - 6)y'&= \sum_{ p = 2 }^{ \infty }\,5(p-1) a_{ p-1 } x^{ p } + \sum_{ j = 0 }^{ \infty }\,-6(j+1) a_{ j+1 } x^{ j }\\ 3y''&= \sum_{ k = 0 }^{ \infty }\,3(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} 3y'' + (5x^2 - 6)y' + (6x - 5)y &= \sum_{ m = 0 }^{ \infty }\,3(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 2 }^{ \infty }\,5(m-1) a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-6(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,6 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-5 a_{ m } x^{ m } \end{align} STEP 4: Collect like terms. \begin{align} &(6a_2 - 6a_1 - 5a_0) + (18a_{3} - 12a_{2} - 5a_{1} + 6a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(3(m+2)(m+1)a_{m+2} - 6(m+1)a_{m+1} + 5(m-1)a_{m-1} - 5a_m + 6a_{m-1})x^m = 0 \end{align} STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$a_2 = \frac{6a_1 + 5a_0}{6}$$ Equating the linear term to 0 we get $$a_3 = \frac{12a_2 + 5a_1 - 6a_0}{18}$$ Finally, equating the general term to 0, we find that for $m \ge 2,$ $$a_{m+2} = \frac{6(m+1)a_{m+1} + 5a_m - (5(m-1)a_m + 6a_m)a_{m-1}}{3(m+2)(m+1)}$$ STEP 6: We know that $a_0 = y(0) = -2$ and $a_1 = y'(0) = 4.$ 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 $$a_2 = \frac{6(4) + 5(-2)}{6} = 7/3$$ From the equation for the linear term we get $$a_3 = \frac{12(7/3) + 5(4) - 6(-2)}{18} = 10/3$$ Finally, using the recurrence equation with $m = 2$ we get $$a_4 = \frac{6(2+1)(10/3) + 5(7/3) - (5(2-1)(7/3) + 6(7/3))(4)}{3(4)(3)} = 83/108$$

So our solution is $$y(x) = -2 + 4x + (7/3)x^2 + (10/3)x^3 + (83/108)x^4 + \cdots$$