Show that the half-life of solutions to (2)—that is, the time required for the solution to decay to one-half of its value—equals\({\bf{(}}\frac{{{\bf{ln2}}}}{{\bf{k}}}{\bf{)}}\).

Given,

\(\begin{aligned}{c}\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = - kp}}\\\int {\frac{{\bf{1}}}{{\bf{p}}}{\bf{dy}}} {\bf{ = }}\int {{\bf{ - kdt}}} \\{\bf{ln}}\left| {\bf{p}} \right|{\bf{ = - kt + c}}\\{\bf{p = c}}{{\bf{e}}^{{\bf{ - kt}}}}\end{aligned}\)

The initial conditions are\({\bf{p(0) = }}{{\bf{p}}_{\bf{o}}}\).

\(\begin{aligned}{c}{{\bf{p}}_{\bf{o}}}{\bf{ = c}}{{\bf{e}}^{{\bf{ - k0}}}}\\{\bf{C = }}{{\bf{p}}_{\bf{o}}}\\{\bf{p = }}{{\bf{p}}_{\bf{o}}}{{\bf{e}}^{{\bf{ - kt}}}}\end{aligned}\)

\(\begin{aligned}{c}{\bf{p(\iota ) = }}\frac{{{{\bf{p}}_{\bf{o}}}}}{{\bf{2}}}\\{{\bf{p}}_{\bf{o}}}{{\bf{e}}^{{\bf{ - k\iota }}}}{\bf{ = }}\frac{{{{\bf{p}}_{\bf{o}}}}}{{\bf{2}}}\\{{\bf{e}}^{{\bf{ - k\iota }}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}\\{\bf{ - k\iota = ln}}\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)\\{\bf{\iota = }}\frac{{{\bf{ln2}}}}{{\bf{k}}}\end{aligned}\)

This is the required result.