Discussion
Let W(x) denote the Wronskian of two linearly independent solutions y1(x) and y2(x) of the differential equation
\[ \begin{aligned} (x-1)\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + xe^{x}y &= 0, \\ x &\gt 1. \end{aligned} \]
If W(2) = 2, then the value of W(5) is
Answer: A
The value is \(W(5) = \frac{1}{8}\).
Write the equation in standard form \(y'' + p(x)y' + q(x)y = 0\) with \(p(x) = \frac{2}{x-1}\). Abel's formula gives the Wronskian:
\[ \begin{aligned} W(x) &= C\exp\left(-\int \frac{2}{x-1}\,dx\right) \\ &= \frac{C}{(x-1)^{2}} \end{aligned} \]
W(2) = C/1 = 2, so C = 2 and \(W(5) = 2/16 = 1/8\).
The term \(xe^{x}y\) plays no part, since Abel's formula uses only the coefficient of \(y'\). Getting 8 comes from using \(\exp(+\int p\,dx)\), a sign error that gives \(W = C(x-1)^{2}\); 1/5 and 5 do not follow from the formula.
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