Discussion
Let x(t), y(t), \(t \in \mathbb{R}\), be two functions satisfying the following system of differential equations:
\(x'(t) = y(t)\),
\(y'(t) = x(t)\),
and \(x(0) = \alpha\), \(y(0) = \beta\), where \(\alpha, \beta\) are real numbers.
Which of the following statements is/are correct?
Answer: A
The correct statement is If \(\alpha = 1, \beta = -1\), then \(|x(t)|+|y(t)| \to 0\).
Differentiating gives \(x'' = x\), so with the initial values:
\[ \begin{aligned} x(t) &= \alpha\cosh t + \beta\sinh t \\ &= \tfrac{\alpha+\beta}{2}e^{t} + \tfrac{\alpha-\beta}{2}e^{-t} \\ y(t) &= \tfrac{\alpha+\beta}{2}e^{t} - \tfrac{\alpha-\beta}{2}e^{-t} \end{aligned} \]
The solution decays only if the growing mode is absent, that is \(\alpha + \beta = 0\). For \(\alpha = 1, \beta = -1\), \(x = e^{-t}\) and \(y = -e^{-t}\), so \(|x|+|y| = 2e^{-t} \to 0\).
In the other three cases \(\alpha + \beta\) is 2, 0.01 and 2.01, all non-zero, so the \(e^{t}\) term grows without bound. Even a tiny perturbation such as 1.01 destroys the decay, because the origin is a saddle point of this system.
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