To solve this problem, we need to find the value of the Wronskian \( W(x) \) at \( x = \frac{1}{2} \). The given differential equation is:
\[(1 + x^2) y'' - x y' + (\cos^2 x) y = 0\]where \( y_1(x) \) and \( y_2(x) \) are linearly independent solutions satisfying the initial conditions:
\[y_1(0) = 3, \quad y_1'(0) = -1, \quad y_2(0) = -5, \quad y_2'(0) = 2.\]The Wronskian \( W(x) \) is defined as:
\[ W(x) = \left| \begin{array}{cc} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \\ \end{array} \right| = y_1(x) y_2'(x) - y_2(x) y_1'(x) \]For linear differential equations of the form:
\[ P(x) y'' + Q(x) y' + R(x) y = 0 \]the Wronskian satisfies the equation:
\[ W(x) = C \, e^{-\int \frac{Q(x)}{P(x)} \, dx} \]where \( C \) is a constant. For our given differential equation, \( P(x) = 1 + x^2 \) and \( Q(x) = -x \). Thus,
\[ \frac{Q(x)}{P(x)} = \frac{-x}{1 + x^2} \]We calculate the integral:
\[ \int \frac{-x}{1 + x^2} \, dx = -\frac{1}{2} \ln(1 + x^2) + C_0 \]Substituting into the Wronskian formula, we get:
\[ W(x) = C \, e^{\frac{1}{2} \ln(1 + x^2)} = C \, \sqrt{1 + x^2} \]Using initial conditions to find \( C \):
\[ W(0) = \left| \begin{array}{cc} 3 & -5 \\ -1 & 2 \\ \end{array} \right| = (3 \cdot 2) - (-5 \cdot -1) = 6 - 5 = 1 \]So,
\[ W(0) = C \, \sqrt{1 + 0^2} = C = 1 \]Thus, the Wronskian becomes:
\[ W(x) = \sqrt{1 + x^2} \]Finally, evaluating \( W\left( \frac{1}{2} \right) \):
\[ W\left( \frac{1}{2} \right) = \sqrt{1 + \left(\frac{1}{2}\right)^2} = \sqrt{1.25} = \frac{\sqrt{5}}{2} \]Therefore, the value of \( W\left( \frac{1}{2} \right) \) is \( \frac{\sqrt{5}}{2} \).