Discussion
The solution \(y(x) = \begin{pmatrix} y_1(x) \\ y_2(x) \end{pmatrix}\) of the initial value problem
\[ \frac{dy}{dx} = \begin{pmatrix} -3 & 4 \\ -2 & 3 \end{pmatrix} y, \quad y(0) = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \]
is equal to
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.
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