Discussion
\(\frac{s}{s^{2} + a^{2}}\) is the Laplace transform of :
Answer: A
\(\frac{s}{s^{2} + a^{2}}\) is the Laplace transform of \(\cos at\).
From the definition, \(\mathcal{L}\{e^{iat}\} = \frac{1}{s - ia} = \frac{s + ia}{s^{2} + a^{2}}\). Its real part, \(\frac{s}{s^{2} + a^{2}}\), is the transform of \(\cos at\), and its imaginary part, \(\frac{a}{s^{2} + a^{2}}\), is the transform of \(\sin at\).
The hyperbolic functions have a minus sign in the denominator: \(\cosh at\) transforms to \(\frac{s}{s^{2} - a^{2}}\) and \(\sinh at\) to \(\frac{a}{s^{2} - a^{2}}\). So sin, sinh and cosh are all ruled out.
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