Discussion
Let \(\delta(t)\) be the unit impulse function defined by \(\delta(x - x_0) = 0\) when \(x \neq x_0\) and
\[ \int_{-\infty}^{\infty} \delta(x - x_0)\, dx = 1. \]
Consider the following initial value problem
\[ 2\frac{d^2 y}{dx^2} + \frac{dy}{dx} + 2y = \delta(x - 5) \]
with \(y(0) = 0\) and \(\frac{dy}{dx}(0) = 0\). Which one of the following is TRUE?
Answer: B
The answer is \(y(10) = \frac{2}{\sqrt{15}} e^{-5/4} \sin\left(\frac{5\sqrt{15}}{4}\right)\).
Take Laplace transforms with zero initial data:
\[ \begin{aligned} (2s^2 + s + 2) Y &= e^{-5s} \\ Y &= \frac{e^{-5s}}{2[(s + \frac{1}{4})^2 + \frac{15}{16}]} \end{aligned} \]
Since \(\frac{1}{(s+a)^2 + b^2}\) inverts to \(\frac{1}{b} e^{-at} \sin bt\), with \(a = \frac{1}{4}\) and \(b = \frac{\sqrt{15}}{4}\), and the factor \(e^{-5s}\) shifts by 5:
\[ \begin{aligned} y(x) = &\frac{2}{\sqrt{15}} e^{-(x-5)/4} \\ &\times \sin \frac{\sqrt{15}(x-5)}{4} \end{aligned} \]
for \(x \gt 5\). At \(x = 10\), \(x - 5 = 5\), which gives the keyed value. The options with \(e^{5/4}\) have the wrong sign in the decay, and those with \(\frac{10}{\sqrt{15}}\) and \(\sin \frac{\sqrt{15}}{4}\) put the 5 in front instead of inside the sine.
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