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

Equating the constant term to 0 we get $$ a_2 = \frac{3a_1 - 2a_0}{-12} $$ Equating the linear term to 0 we get $$ a_3 = \frac{6a_2 - 2a_1 - 5a_0}{-36} $$ Finally, equating the general term to 0, we find that for $ m \ge 2,$ $$ a_{m+2} = \frac{3(m+1)a_{m+1} - 2a_m - (5(m-1)a_m + 5a_m)a_{m-1}}{-6(m+2)(m+1)} $$ STEP 6: We know that $ a_0 = y(0) = 1 $ 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 $$ a_2 = \frac{3(-2) - 2(1)}{-12} = 2/3 $$ From the equation for the linear term we get $$ a_3 = \frac{6(2/3) - 2(-2) - 5(1)}{-36} = -1/12 $$ Finally, using the recurrence equation with $ m = 2 $ we get $$ a_4 = \frac{3(2+1)(-1/12) - 2(2/3) - (5(2-1)(2/3) + 5(2/3))(-2)}{-6(4)(3)} = -215/864 $$

So our solution is $$ y(x) = 1 - 2x + (2/3)x^2 - (1/12)x^3 - (215/864)x^4 + \cdots $$

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