Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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\({\bf{(}}\frac{{{\bf{ln2}}}}{{\bf{k}}}{\bf{)}}\).

 

Step-by-Step Solution

Answer

For the result find the value of \({\bf{p(\iota ) = }}\frac{{{{\bf{p}}_{\bf{o}}}}}{{\bf{2}}}\)with initial conditions.

Step 1: Use the solution of (2)

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\({\bf{p(0) = }}{{\bf{p}}_{\bf{o}}}\).

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

Step 2: Find the result

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

 

This is the required result.