In this problem you will solve the differential equation $$(x^2 + 4) y'' + 7 x y'-y =0.$$

(1) By analyzing the singular points of the differential equation, we know that a series solution of the form $y = \sum_{k=0}^{\infty}c_k\ x^k$ for the differential equation will converge at least on the interval .

(2) Substituting $y = \sum_{k=0}^{\infty}c_k\ x^k$ into $(x^2 + 4) y'' + 7 x y' - y =0$, you get that

$\Big($

c

$-$

c

$\Big)+\Big($

c

$+$

c

$\Big)x+$

$\hskip 3pt\infty$ $\displaystyle\sum$ $n$$=$$2$

$\Bigg\lbrack$

c

$+$

c

$\Bigg\rbrack x^n = 0$

The subscripts on the $c$'s should be increasing and numbers or in terms of $n$.

(3) In this step we will use the equation above to solve for some of the terms in the series and find the recurrence relation.
(a) From the constant term in the series above, we know that

(b) From the coefficient of $x$ in the series above, we know that

(c) From the series above, we find that the recurrence relation is

(4) The general solution to $(x^2 + 4) y'' + 7 x y' - y=0$ converges at least on and is
$y = c_0\Big($ $+$ $x^2+$ $x^4+$ $x^6+\cdots\Big) + c_1\Big($ $x+$ $x^3+$ $x^5+$ $x^7+\cdots\Big)$