Let \( \alpha \) and \( \beta \) be the roots of the indicial equation obtained in the method of finding the Frobenius series solution to the differential equation:
\[ 2x^2 \frac{d^2y}{dx^2} - x\frac{dy}{dx} + (1 - x^2) y = 0. \]
Which of the following statements is/are TRUE?
Show Hint
Substitute y = x^r into only the lowest-power terms of the equation to get a quadratic in r, then solve it and compare the two roots.
Step 1: Convert to the standard form to read off $p_0$ and $q_0$.
Divide the whole equation $2x^2y'' - xy' + (1-x^2)y = 0$ by $2x^2$ to get $y'' + p(x)y' + q(x)y = 0$:
$$y'' - \frac{1}{2x}y' + \frac{1-x^2}{2x^2}y = 0$$
Here $p(x) = -\dfrac{1}{2x}$ and $q(x) = \dfrac{1-x^2}{2x^2}$.
Step 2: Find $p_0$ and $q_0$ at the regular singular point $x=0$.
Since $x=0$ is a regular singular point, the limits $p_0 = \lim_{x\to 0} x\,p(x)$ and $q_0 = \lim_{x\to 0} x^2 q(x)$ exist:
$$p_0 = \lim_{x\to 0} x \left(-\frac{1}{2x}\right) = -\frac{1}{2}, \qquad q_0 = \lim_{x\to 0} x^2 \cdot \frac{1-x^2}{2x^2} = \lim_{x\to 0}\frac{1-x^2}{2} = \frac{1}{2}$$
Step 3: Write and solve the indicial equation using $r(r-1) + p_0 r + q_0 = 0$.
$$r(r-1) - \frac{1}{2}r + \frac{1}{2} = 0$$
Multiply through by 2 to clear the fraction:
$$2r^2 - 2r - r + 1 = 0 \implies 2r^2 - 3r + 1 = 0$$
Using the quadratic formula, $r = \dfrac{3 \pm \sqrt{9-8}}{4} = \dfrac{3\pm1}{4}$, so $r = 1$ or $r = \dfrac12$.
Step 4: Compare the roots against each option.
Let $\alpha = 1$, $\beta = \tfrac12$. Their difference is $\alpha - \beta = \tfrac12$, which is not a whole number.
So $\alpha$ and $\beta$ are unequal (this confirms option A and rules out option D), and $\alpha - \beta$ is not an integer (this confirms option C and rules out option B).
Final Answer:
$$\boxed{\alpha \neq \beta \text{ and } \alpha - \beta \notin \mathbb{Z}}$$
