Question

Consider the ordinary differential equation (ODE) \[ 4 (\ln x) y'' + 3 y' + y = 0, \quad x>1. \] If \(r_1\) and \(r_2\) are the roots of the indicial equation of the above ODE at the regular singular point \(x=1\), then \(|r_1 - r_2|\) is equal to ................. (rounded off to 2 decimal places).

Hint:

When dealing with a regular singular point \(x_0\), remember the standard limits for \(p_0\) and \(q_0\). For limits involving \(\ln x\) near \(x=1\), the substitution \(t = x-1\) and the Taylor expansion \(\ln(1+t) \approx t\) can be very useful. For example, \(\lim_{x \to 1} \frac{x-1}{\ln x} = \lim_{t \to 0} \frac{t}{\ln(1+t)} = \lim_{t \to 0} \frac{t}{t} = 1\).

Answer 1

Step 1: Understanding the Concept:
This problem requires finding the roots of the indicial equation for a second-order linear ODE at a regular singular point. The method of Frobenius is used for such points. The indicial equation is a quadratic equation whose roots determine the form of the series solution.
Step 2: Key Formula or Approach:
For an ODE of the form \(P(x)y'' + Q(x)y' + R(x)y = 0\), a point \(x_0\) is a regular singular point if \(p(x) = (x-x_0)\frac{Q(x)}{P(x)}\) and \(q(x) = (x-x_0)^2\frac{R(x)}{P(x)}\) are analytic at \(x_0\).
The indicial equation is given by: \[ r(r-1) + p_0 r + q_0 = 0 \] where \(p_0 = \lim_{x \to x_0} p(x)\) and \(q_0 = \lim_{x \to x_0} q(x)\).
Step 3: Detailed Explanation or Calculation:
The given ODE is \(4 (\ln x) y'' + 3 y' + y = 0\). The singular point is \(x=1\), since \(\ln(1) = 0\). Let's find \(p_0\) and \(q_0\) for \(x_0 = 1\). Here, \(P(x) = 4 \ln x\), \(Q(x) = 3\), and \(R(x) = 1\). \[ p_0 = \lim_{x \to 1} (x-1) \frac{Q(x)}{P(x)} = \lim_{x \to 1} (x-1) \frac{3}{4 \ln x} \] This is an indeterminate form \(\frac{0}{0}\). We use L'Hôpital's Rule: \[ p_0 = \lim_{x \to 1} \frac{\frac{d}{dx}(3(x-1))}{\frac{d}{dx}(4 \ln x)} = \lim_{x \to 1} \frac{3}{4/x} = \frac{3}{4/1} = \frac{3}{4} \] Now, let's find \(q_0\): \[ q_0 = \lim_{x \to 1} (x-1)^2 \frac{R(x)}{P(x)} = \lim_{x \to 1} (x-1)^2 \frac{1}{4 \ln x} \] We can write this as: \[ q_0 = \lim_{x \to 1} \frac{x-1}{4 \ln x} . (x-1) \] We already know from the calculation of \(p_0\) that \(\lim_{x \to 1} \frac{x-1}{\ln x} = 1\). \[ q_0 = \frac{1}{4} \left( \lim_{x \to 1} \frac{x-1}{\ln x} \right) \left( \lim_{x \to 1} (x-1) \right) = \frac{1}{4} . (1) . (0) = 0 \] Now, we form the indicial equation: \[ r(r-1) + p_0 r + q_0 = 0 \] \[ r(r-1) + \frac{3}{4}r + 0 = 0 \] \[ r^2 - r + \frac{3}{4}r = 0 \] \[ r^2 - \frac{1}{4}r = 0 \] \[ r \left( r - \frac{1}{4} \right) = 0 \] The roots of the indicial equation are \(r_1 = 0\) and \(r_2 = \frac{1}{4}\).
Step 4: Final Answer:
The question asks for the absolute difference between the roots, \(|r_1 - r_2|\). \[ |r_1 - r_2| = \left| 0 - \frac{1}{4} \right| = \frac{1}{4} = 0.25 \] The value is 0.25.
Step 5: Why This is Correct:
The regular singular point is correctly identified. The limits \(p_0\) and \(q_0\) are calculated accurately using L'Hôpital's rule. The indicial equation is formulated and solved correctly. The final calculation of the absolute difference is straightforward and correct.