Maths Question:
$\text{If }y=A{{e}^{mx}}+B{{e}^{-mx}},\text{ where }A,B,\text{ and }M\text{ are constant Show that }\frac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}y$
Maths Solution:
$\begin{align} & y=A{{e}^{mx}}+B{{e}^{-mx}} \\ & \frac{dy}{dx}=Am{{e}^{mx}}-Bm{{e}^{-mx}} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=A{{m}^{2}}{{e}^{mx}}+B{{m}^{2}}{{e}^{-mx}} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}(A{{e}^{mx}}+B{{e}^{-mx}}) \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}y \\\end{align}$
University mathstopic: