Systems of ODEs and Stability Questions and Answers

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.

Answer: A

The origin is stable for any such matrix M.

If M is skew-symmetric, then

\[ \frac{d}{dt}\|X\|^{2} = X^{T}(M + M^{T})X = 0 \]

so every solution keeps a constant length. A solution that starts within distance \(\varepsilon\) of the origin stays within \(\varepsilon\) for all time, which is Lyapunov stability. Equivalently, the eigenvalues of a real skew-symmetric matrix are purely imaginary (or zero) and the matrix is diagonalizable, so the solutions are bounded rotations.

The rank does not matter: a zero eigenvalue just gives constant components. The stability is not asymptotic, since solutions do not decay, but the question asks only for stability, so the rank-restricted options are wrong.

Answer: D

The solution is \(3 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{x}\) - \(\begin{pmatrix} 2 \\ 1 \end{pmatrix} e^{-x}\).

The matrix has trace 0 and determinant \(-9 + 8 = -1\), so its eigenvalues are \(\pm 1\). For \(\lambda = 1\), \((A - I)v = 0\) gives \(v = (1, 1)^{\top}\); for \(\lambda = -1\), \((A + I)v = 0\) gives \(v = (2, 1)^{\top}\). So

\[ y = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{x} + c_2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} e^{-x} \]

and \(y(0) = (1, 2)^{\top}\) gives

\[ \begin{aligned} c_1 + 2c_2 &= 1 \\ c_1 + c_2 &= 2 \\ c_2 = -1, \ c_1 &= 3 \end{aligned} \]

The option with \(+\) gives \(y(0) = (5, 4)^{\top}\), the one with \(-3\) gives \((-1, -2)^{\top}\), and the one with \(2(2,1)^{\top}\) gives \((7, 5)^{\top}\); none matches the initial condition.

Answer: C

The solutions satisfy \(3y + z = ce^{3x}\).

Differentiate the combination \(3y + z\):

\[ \begin{aligned} \frac{d}{dx}(3y + z) &= 3(2y + z) + 3y \\ &= 9y + 3z = 3(3y + z) \end{aligned} \]

so \(3y + z = ce^{3x}\). The matrix \(\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}\) has eigenvalues 3 and \(-1\), and the other combination is \(y - z\): \(\frac{d}{dx}(y - z) = 2y + z - 3y = -(y - z)\), so \(y - z = ce^{-x}\).

So option A pairs \(y - z\) with the wrong exponential, option D pairs \(3y + z\) with the wrong one, and \(y + z\) is not an eigen-combination at all.