Question:
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?
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.
Updated On: Jul 21, 2026
- \( \alpha \neq \beta \)
- \( \alpha - \beta \in \mathbb{Z} \)
- \( \alpha - \beta \notin \mathbb{Z} \)
- \( \alpha = \beta \)
Show Solution
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The Correct Option is A, C
Solution and Explanation
Step 1: Write the equation near the singular point.
The given equation is \( 2x^2 y'' - x y' + (1 - x^2) y = 0 \), where \(y'' = \dfrac{d^2y}{dx^2}\) and \(y' = \dfrac{dy}{dx}\).
Here \(x = 0\) is a regular singular point of the equation, so we look for a Frobenius series solution of the form \(y = \sum_{n=0}^{\infty} a_n x^{n+r}\) with \(a_0 \neq 0\).
Step 2: Substitute the lowest power term to get the indicial equation.
Put \(y = x^r\) and keep only the terms that come from the lowest power of \(x\), since these fix the indicial equation.
\(y' = r x^{r-1}\) and \(y'' = r(r-1)x^{r-2}\).
So \(2x^2 y'' = 2r(r-1)x^r\), \(-xy' = -r x^r\), and the leading part of \((1-x^2)y\) is \(1 \cdot x^r\) (the \(-x^2 y\) term shifts the power up by 2 and does not enter the indicial equation).
Adding the coefficients of \(x^r\) and setting the sum to zero gives the indicial equation:
\[ 2r(r-1) - r + 1 = 0 \]
Step 3: Solve the indicial equation.
Expand:
\[ 2r^2 - 2r - r + 1 = 0 \implies 2r^2 - 3r + 1 = 0 \] Factor:
\[ (2r - 1)(r - 1) = 0 \] So the roots are \(r = 1\) and \(r = \dfrac{1}{2}\).
Take \(\alpha = 1\) and \(\beta = \dfrac{1}{2}\) (which root we call \(\alpha\) and which we call \(\beta\) does not change the size of the difference).
Step 4: Check each option.
(A) \(\alpha \neq \beta\): since \(1 \neq \dfrac{1}{2}\), this is TRUE.
(B) \(\alpha - \beta \in \mathbb{Z}\): \(\alpha - \beta = 1 - \dfrac{1}{2} = \dfrac{1}{2}\), which is not an integer, so this is FALSE.
(C) \(\alpha - \beta \notin \mathbb{Z}\): since \(\dfrac{1}{2}\) is not an integer, this is TRUE.
(D) \(\alpha = \beta\): the two roots are different, so this is FALSE.
Final Answer:
The roots of the indicial equation are \(1\) and \(\dfrac{1}{2}\), so they are distinct and their difference is not a whole number.
\[ \boxed{\text{(A) and (C) are TRUE}} \]
The given equation is \( 2x^2 y'' - x y' + (1 - x^2) y = 0 \), where \(y'' = \dfrac{d^2y}{dx^2}\) and \(y' = \dfrac{dy}{dx}\).
Here \(x = 0\) is a regular singular point of the equation, so we look for a Frobenius series solution of the form \(y = \sum_{n=0}^{\infty} a_n x^{n+r}\) with \(a_0 \neq 0\).
Step 2: Substitute the lowest power term to get the indicial equation.
Put \(y = x^r\) and keep only the terms that come from the lowest power of \(x\), since these fix the indicial equation.
\(y' = r x^{r-1}\) and \(y'' = r(r-1)x^{r-2}\).
So \(2x^2 y'' = 2r(r-1)x^r\), \(-xy' = -r x^r\), and the leading part of \((1-x^2)y\) is \(1 \cdot x^r\) (the \(-x^2 y\) term shifts the power up by 2 and does not enter the indicial equation).
Adding the coefficients of \(x^r\) and setting the sum to zero gives the indicial equation:
\[ 2r(r-1) - r + 1 = 0 \]
Step 3: Solve the indicial equation.
Expand:
\[ 2r^2 - 2r - r + 1 = 0 \implies 2r^2 - 3r + 1 = 0 \] Factor:
\[ (2r - 1)(r - 1) = 0 \] So the roots are \(r = 1\) and \(r = \dfrac{1}{2}\).
Take \(\alpha = 1\) and \(\beta = \dfrac{1}{2}\) (which root we call \(\alpha\) and which we call \(\beta\) does not change the size of the difference).
Step 4: Check each option.
(A) \(\alpha \neq \beta\): since \(1 \neq \dfrac{1}{2}\), this is TRUE.
(B) \(\alpha - \beta \in \mathbb{Z}\): \(\alpha - \beta = 1 - \dfrac{1}{2} = \dfrac{1}{2}\), which is not an integer, so this is FALSE.
(C) \(\alpha - \beta \notin \mathbb{Z}\): since \(\dfrac{1}{2}\) is not an integer, this is TRUE.
(D) \(\alpha = \beta\): the two roots are different, so this is FALSE.
Final Answer:
The roots of the indicial equation are \(1\) and \(\dfrac{1}{2}\), so they are distinct and their difference is not a whole number.
\[ \boxed{\text{(A) and (C) are TRUE}} \]
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