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### Series Solutions

Find the series solution (up to the $x^4$ term) for the equation $$-5y'' + (-6x - 7)y' + (3x + 3)y = 0,\qquad y(0) = -4, y'(0) = -3$$

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 + 3)y &= \sum_{ n = 0 }^{ \infty }\,3 a_{ n } x^{ n+1 } + \sum_{ n = 0 }^{ \infty }\,3 a_{ n } x^{ n }\\ (-6x - 7)y' &= \sum_{ n = 1 }^{ \infty }\,-6n a_{ n } x^{ n } + \sum_{ n = 1 }^{ \infty }\,-7n a_{ n } x^{ n-1 }\\ -5y'' &= \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 $p = n+1$ 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} (3x + 3)y&= \sum_{ p = 1 }^{ \infty }\,3 a_{ p-1 } x^{ p } + \sum_{ n = 0 }^{ \infty }\,3 a_{ n } x^{ n }\\ (-6x - 7)y'&= \sum_{ n = 1 }^{ \infty }\,-6n a_{ n } x^{ n } + \sum_{ j = 0 }^{ \infty }\,-7(j+1) a_{ j+1 } x^{ j }\\ -5y''&= \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} -5y'' + (-6x - 7)y' + (3x + 3)y &= \sum_{ m = 0 }^{ \infty }\,-5(m+2)(m+1) a_{ m+2 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,-6m a_{ m } x^{ m } + \sum_{ m = 0 }^{ \infty }\,-7(m+1) a_{ m+1 } x^{ m } \\ &+ \sum_{ m = 1 }^{ \infty }\,3 a_{ m-1 } x^{ m } + \sum_{ m = 0 }^{ \infty }\,3 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 - 7a_1 + 3a_0) + (-30a_{3} - 14a_{2} - 6a_{1} + 3a_{1} + 3a_{0})x \\ &+ \sum_{ m = 2 }^{ \infty }\,(-5(m+2)(m+1)a_{m+2} - 7(m+1)a_{m+1} - 6ma_m + 3a_m + 3a_{m-1})x^m = 0 \end{align} STEP 5: Equate coefficients to 0.

Equating the constant term to 0 we get $$a_2 = \frac{7a_1 - 3a_0}{-10}$$ Equating the linear term to 0 we get $$a_3 = \frac{14a_2 + 3a_1 - 3a_0}{-30}$$ Finally, equating the general term to 0, we find that for $m \ge 2,$ $$a_{m+2} = \frac{7(m+1)a_{m+1} - (-6m + 3)a_m - 3a_{m-1}}{-5(m+2)(m+1)}$$ STEP 6: We know that $a_0 = y(0) = -4$ 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 $$a_2 = \frac{7(-3) - 3(-4)}{-10} = 9/10$$ From the equation for the linear term we get $$a_3 = \frac{14(9/10) + 3(-3) - 3(-4)}{-30} = -13/25$$ Finally, using the recurrence equation with $m = 2$ we get $$a_4 = \frac{7(2+1)(-13/25) - (-62 + 3)(9/10) - 3(-3)}{-5(4)(3)} = -103/1000$$

So our solution is $$y(x) = -4 - 3x + (9/10)x^2 - (13/25)x^3 - (103/1000)x^4 + \cdots$$