Discussion
The sum of the roots of the indicial equation at \(x = 0\) of the differential equation
\[ \begin{aligned} & x^{3}\frac{d^{2}y}{dx^{2}} + (x \sin x)\frac{dy}{dx} \\ & \quad - (\tan x)y = 0, \quad x \gt 0, \end{aligned} \]
is
Answer: A
The sum of the roots is 0.
In standard form the equation is \(y'' + \frac{\sin x}{x^{2}}y' - \frac{\tan x}{x^{3}}y = 0\). At x = 0,
\[ \begin{aligned} p_{0} &= \lim_{x \to 0} x \cdot \frac{\sin x}{x^{2}} = 1 \\ q_{0} &= \lim_{x \to 0} x^{2} \cdot \left(-\frac{\tan x}{x^{3}}\right) = -1 \end{aligned} \]
so x = 0 is a regular singular point. The indicial equation \(r(r - 1) + p_{0}r + q_{0} = 0\) becomes \(r^{2} - 1 = 0\), with roots 1 and -1, whose sum is 0. (In general the sum is \(1 - p_{0}\).) Taking \(p_{0} = 0\) would give a sum of 1, and neither 2 nor -2 can arise from roots of \(r^{2} = 1\).
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