Discussion
The solutions of the differential equations
\[ \frac{dy}{dx} = 2y + z \;\text{ and }\; \frac{dz}{dx} = 3y \]
satisfy the relation :
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.
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