Discussion
Given that the Laplace transforms of \(J_{0}(x)\), \(J_{0}'(x)\) and \(J_{0}''(x)\) exist, where \(J_{0}(x)\) is the Bessel function. Let \(Y = Y(s)\) be the Laplace transform of the Bessel function \(J_{0}(x)\). Then, which one of the following is TRUE?
Answer: D
The correct equation is \(\dfrac{dY}{ds} + \dfrac{sY}{s^{2} + 1} = 0\).
\(J_{0}\) satisfies Bessel's equation \(xy'' + y' + xy = 0\) with \(y(0) = 1\), \(y'(0) = 0\). Using \(\mathcal{L}\{xf\} = -\frac{d}{ds}\mathcal{L}\{f\}\):
\[ \begin{aligned} \mathcal{L}\{xy''\} &= -\tfrac{d}{ds}(s^{2}Y - s) = -2sY - s^{2}Y' + 1 \\ \mathcal{L}\{y'\} &= sY - 1, \quad \mathcal{L}\{xy\} = -Y' \\ 0 &= -(s^{2} + 1)Y' - sY \end{aligned} \]
Dividing by \(-(s^{2} + 1)\) gives the equation. Its solution, \(Y = 1/\sqrt{s^{2} + 1}\), is the known transform of \(J_{0}\). The options with \(2sY\) come from dropping the \(sY\) of the \(y'\) term, and the minus signs reverse the result.
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