Discussion
The largest interval in which the initial value problem
\[ \begin{aligned} & e^{x}\frac{d^{2}y}{dx^{2}} + \frac{1}{(x-5)}\frac{dy}{dx} \\ & \quad + \left(\sqrt{x}\right)y = \ln(x), \end{aligned} \]
\(y(1) = 0\) and \(\frac{dy}{dx}(1) = 1\),
has a unique solution is
Answer: D
The largest interval is \((0, 5)\).
Divide by \(e^{x}\) to write the equation as \(y'' + p(x)y' + q(x)y = g(x)\) with
\[ \begin{aligned} p &= \frac{e^{-x}}{x - 5}, \quad q = e^{-x}\sqrt{x} \\ g &= e^{-x}\ln x \end{aligned} \]
The existence and uniqueness theorem guarantees a unique solution on the largest open interval containing the initial point x = 1 on which p, q and g are all continuous. \(\sqrt{x}\) and \(\ln x\) need \(x \gt 0\), and p blows up at x = 5. The interval containing 1 is therefore (0, 5).
\((-\infty, \infty)\) and \((-5, 5)\) include negative x, where \(\ln x\) is undefined, and \((0, \infty)\) crosses the singularity at x = 5.
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