Discussion
If \(y = \sum_{m=0}^{\infty} c_{m} x^{r+m}\) is assumed to be a solution of the differential equation
\[ x^{2}y'' - xy' - 3(1 + x^{2})y = 0, \]
then the values of \(r\) are
Answer: B
The exponents are -1 and 3.
Substituting \(y = \sum c_{m}x^{r+m}\) and collecting the lowest power \(x^{r}\) gives the indicial equation from the terms \(x^{2}y''\), \(-xy'\) and \(-3y\) (the \(-3x^{2}y\) term raises the power by two and does not contribute):
\[ \begin{aligned} r(r-1) - r - 3 &= 0 \\ r^{2} - 2r - 3 &= 0 \\ (r - 3)(r + 1) &= 0 \end{aligned} \]
So \(r = 3\) or \(r = -1\). The other pairs come from sign errors: '1 and 3' and '1 and -3' change the sign of one root, and '-1 and -3' solves \(r^{2} + 4r + 3 = 0\), which is not the indicial equation here.
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