Question

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?

• A. If \(\alpha = 1, \beta = -1\), then \(|x(t)| + |y(t)| \to 0\) as \(t \to \infty\)
• B. If \(\alpha = 1, \beta = 1\), then \(|x(t)| + |y(t)| \to 0\) as \(t \to \infty\)
• C. If \(\alpha = 1.01, \beta = -1\), then \(|x(t)| + |y(t)| \to 0\) as \(t \to \infty\)
• D. If \(\alpha = 1, \beta = 1.01\), then \(|x(t)| + |y(t)| \to 0\) as \(t \to \infty\)

Hint:

For a linear system with constant coefficients, the long-term behavior is dominated by the term \(e^{\lambda t}\) corresponding to the eigenvalue \(\lambda\) with the largest real part. For the solution to decay to zero, all eigenvalues must have negative real parts. Here, the eigenvalues of the system matrix \(\begin{pmatrix} 0 & 1
1 & 0 \end{pmatrix}\) are \(\pm 1\). The solution will decay only if the initial condition lies entirely in the eigenspace of the negative eigenvalue (\(\lambda=-1\)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a system of first-order linear homogeneous differential equations. We need to find the general solution and then use the initial conditions to find the particular solution. Finally, we must analyze the long-term behavior (\(t \to \infty\)) of the solution for different initial conditions.
Step 2: Key Formula or Approach:
1. Convert the system into a single second-order ODE.
2. Find the general solution of the second-order ODE.
3. Use the general solution to find the solution for the other variable.
4. Apply the initial conditions to determine the constants.
5. Analyze the limit of \(|x(t)| + |y(t)|\) as \(t \to \infty\).
Step 3: Detailed Calculation:
From \(x'(t) = y(t)\), we can differentiate to get \(x''(t) = y'(t)\).
Substituting \(y'(t) = x(t)\) from the second equation gives \(x''(t) = x(t)\), or \(x'' - x = 0\).
The characteristic equation is \(r^2 - 1 = 0\), which has roots \(r_1 = 1\) and \(r_2 = -1\). The general solution for \(x(t)\) is: \[ x(t) = c_1 e^t + c_2 e^{-t} \] Then \(y(t)\) is the derivative of \(x(t)\): \[ y(t) = x'(t) = c_1 e^t - c_2 e^{-t} \] Now we apply the initial conditions \(x(0) = \alpha\) and \(y(0) = \beta\): \[ x(0) = c_1 e^0 + c_2 e^0 = c_1 + c_2 = \alpha \] \[ y(0) = c_1 e^0 - c_2 e^0 = c_1 - c_2 = \beta \] We solve this system for \(c_1\) and \(c_2\). Adding the two equations gives \(2c_1 = \alpha + \beta\), so \(c_1 = \frac{\alpha + \beta}{2}\).
Subtracting the second from the first gives \(2c_2 = \alpha - \beta\), so \(c_2 = \frac{\alpha - \beta}{2}\).
The particular solution is: \[ x(t) = \frac{\alpha+\beta}{2}e^t + \frac{\alpha-\beta}{2}e^{-t} \] \[ y(t) = \frac{\alpha+\beta}{2}e^t - \frac{\alpha-\beta}{2}e^{-t} \] We want to find the condition for which \(|x(t)| + |y(t)| \to 0\) as \(t \to \infty\). As \(t \to \infty\), \(e^{-t} \to 0\), but \(e^t \to \infty\). For the solution to go to zero, the terms involving \(e^t\) must be eliminated. This requires their coefficient to be zero: \[ \frac{\alpha + \beta}{2} = 0 \implies \alpha + \beta = 0 \] If this condition holds, then \(x(t) = \frac{\alpha-\beta}{2}e^{-t}\) and \(y(t) = -\frac{\alpha-\beta}{2}e^{-t}\). Both terms go to 0 as \(t \to \infty\), and so does their sum of absolute values.
Now we check the options based on the condition \(\alpha + \beta = 0\):
(A) \(\alpha = 1, \beta = -1\): \(\alpha + \beta = 1 + (-1) = 0\). The condition is satisfied. So, \(|x(t)| + |y(t)| \to 0\). (A) is TRUE.
(B) \(\alpha = 1, \beta = 1\): \(\alpha + \beta = 1 + 1 = 2 \neq 0\). The limit goes to \(\infty\). (B) is FALSE.
(C) \(\alpha = 1.01, \beta = -1\): \(\alpha + \beta = 1.01 - 1 = 0.01 \neq 0\). The limit goes to \(\infty\). (C) is FALSE.
(D) \(\alpha = 1, \beta = 1.01\): \(\alpha + \beta = 1 + 1.01 = 2.01 \neq 0\). The limit goes to \(\infty\). (D) is FALSE.
Step 4: Final Answer:
The only correct statement is (A).