Discussion
\(y = ae^{mx} + be^{-mx}\) is the solution of the differential equation
Answer: D
It solves \(\frac{d^{2}y}{dx^{2}} - m^{2}y = 0\).
Differentiate twice:
\[ \begin{aligned} y' &= ame^{mx} - bme^{-mx} \\ y'' &= am^{2}e^{mx} + bm^{2}e^{-mx} = m^{2}y \end{aligned} \]
So \(y'' - m^{2}y = 0\). The two arbitrary constants a and b show that the equation is of second order, which rules out both first-order options: \(y' - my = 0\) is solved by \(ae^{mx}\) alone and \(y' + my = 0\) by \(be^{-mx}\) alone. \(y'' + m^{2}y = 0\) has the oscillating solutions cos mx and sin mx, not exponentials.
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