Discussion
Let \(y_{1}(x)\) and \(y_{2}(x)\) be the two linearly independent solutions of the differential equation
\[ (1 + x^{2})y'' - xy' + (\cos^{2} x)y = 0, \]
satisfying the initial conditions
\[ \begin{aligned} &y_{1}(0) = 3,\ y_{1}'(0) = -1 \ \text{and} \\ &y_{2}(0) = -5,\ y_{2}'(0) = 2. \end{aligned} \]
Define
\[ W(x) = \begin{vmatrix} y_{1}(x) & y_{2}(x) \\ y_{1}'(x) & y_{2}'(x) \end{vmatrix}. \]
Then, the value of \(W\left(\frac{1}{2}\right)\) is
Answer: B
\(W(1/2) = \dfrac{\sqrt{5}}{2}\).
Abel's formula: for \(y'' + p(x)y' + q(x)y = 0\), \(W(x) = W(0)\exp\left(-\int_{0}^{x} p\right)\). Dividing by \(1 + x^{2}\), \(p(x) = -x/(1 + x^{2})\), so
\[ \begin{aligned} W(x) &= W(0)\exp\left(\int_{0}^{x}\frac{s}{1 + s^{2}}ds\right) = W(0)\sqrt{1 + x^{2}} \\ W(0) &= (3)(2) - (-5)(-1) = 1 \\ W(\tfrac{1}{2}) &= \sqrt{1 + \tfrac{1}{4}} = \frac{\sqrt{5}}{2} \end{aligned} \]
The coefficient of y plays no part in Abel's formula. \(2/\sqrt{5}\) is the reciprocal, from a sign slip in the exponent, and \(\sqrt{5}/4\) halves the answer.
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