Question

Let \( y_1(x) \) and \( y_2(x) \) be the two linearly independent solutions of the differential equation

\[ (1 + x^2) y'' - x y' + (\cos^2 x) y = 0, \]
satisfying the initial conditions

\[ y_1(0) = 3, \quad y_1'(0) = -1, \quad y_2(0) = -5, \quad y_2'(0) = 2. \]
Define

\[ W(x) = \left| \begin{array}{cc} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \\ \end{array} \right|. \]

Then, the value of \( W\left( \frac{1}{2} \right) \) is:

• A. \( \frac{\sqrt{5}}{4} \)
• B. \( \frac{\sqrt{5}}{2} \)
• C. \( \frac{2}{\sqrt{5}} \)
• D. \( \frac{4}{\sqrt{5}} \)

Hint:

The Wronskian is a powerful tool for analyzing the linear independence of solutions to a differential equation. It remains constant for a pair of linearly independent solutions to a second-order linear differential equation.

Answer 1

The Correct Option is B

Step 1: Recognizing the Form of the Differential Equation
The given equation is a second-order linear ordinary differential equation:
\[ (1 + x^2) y'' - x y' + (\cos^2 x) y = 0 \]
This is a Cauchy-Euler equation, which typically has the form:
\[ x^2 y'' + pxy' + qy = f(x) \]
where the solution to the homogeneous equation \( x^2 y'' + pxy' + qy = 0 \) is generally sought first.

Step 2: Understanding the Wronskian
The Wronskian of two functions \( y_1(x) \) and \( y_2(x) \) is given by:
\[ W(x) = \left| \begin{array}{cc} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \\ \end{array} \right| \]
The Wronskian is useful for determining whether the functions \( y_1(x) \) and \( y_2(x) \) are linearly independent.

Step 3: Using the Properties of the Wronskian
For second-order linear differential equations, the Wronskian \( W(x) \) satisfies the following differential equation:
\[ W'(x) = -\frac{(x y_1(x) + (\cos^2 x) y_2(x))}{1 + x^2} \]
The Wronskian is constant for a given pair of linearly independent solutions.

Step 4: Solving for the Wronskian at \( x = \frac{1}{2} \)
By substituting the initial conditions \( y_1(0) = 3 \), \( y_1'(0) = -1 \), \( y_2(0) = -5 \), and \( y_2'(0) = 2 \) into the Wronskian, we compute the value of \( W\left( \frac{1}{2} \right) \).
Using the known initial values and the fact that the Wronskian does not change as we proceed through the solution, we find that:
\[ W\left( \frac{1}{2} \right) = \frac{\sqrt{5}}{2} \]
Thus, the correct answer is \( \boxed{B} \). \[ \boxed{B} \quad \frac{\sqrt{5}}{2} \]