Laplace Transforms Questions and Answers

Answer: B

The transform is \(3/(s - 2)^{2}\).

Transform the equation with \(Y = L\{y\}\), \(y(0) = 0\), \(y'(0) = 3\):

\[ \begin{aligned} (s^{2}Y - 3) - 4sY + 4Y &= 0 \\ Y(s^{2} - 4s + 4) &= 3 \\ Y &= \frac{3}{(s - 2)^{2}} \end{aligned} \]

So \(y = 3te^{2t}\), which satisfies both initial conditions. The \((s + 2)^{2}\) options come from a sign slip in the characteristic polynomial, and 3.5 has no source in the data.

Answer: C

The true relation is F{f(t)} = L{g(t)}, with the transforms evaluated at matching arguments.

Taking the Fourier transform of \(f\) with kernel \(e^{-iyt}\):

\[ \begin{aligned} \mathcal{F}\{f\}(y) &= \int_{0}^{\infty} e^{-iyt}e^{-xt}g(t)\, dt \\ &= \int_{0}^{\infty} e^{-(x + iy)t}g(t)\, dt = \mathcal{L}\{g\}(x + iy) \end{aligned} \]

So the Fourier transform of \(f\) is the Laplace transform of \(g\) at \(s = x + iy\); this is how the Laplace transform is obtained from the Fourier transform. No factor \(e^{s}\) or \(s\) appears, which rules out A and B, and D has the two transforms the wrong way round.