Extend the radioactive decay problem (Example 2) one more stage, that is, let ${\mathit{\lambda }}_{\mathbf{3}}$be the decay constant of polonium and find how much polonium there is at time t.

A radioactive decay problem where the Radium decays to Radon which decays to Polonium. The decay constant of polonium is given to be${\lambda }_3$.

For a linear differential equation$y\text{'}+Py=Q$,

$I=\int Pdx$ …… (3.4)

And the general solution is,

$\begin{array}{r}y{e}^I=\int Q{e}^{I}dx+c_{,}or\\ y=e^{-I}\int Q{e}^{I}dx+c{e}^{-I},\end{array}\}\mathbf{\text{where}}\mathit{I}\mathbf{=}\mathbf{\int }\mathit{P}\mathit{d}\mathit{x}$ ……(3.9)

Consider the decay of Polonium.

The rate of change is the difference,

That is,$\frac{d{N}_3}{dt}+{\lambda }_3{N}_3=\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)$

Hence, the equation will be in the form, $y\text{'}+Py=Q$.

By equation (3.4),

$$\begin{gathered} I=\int {\lambda }_{3}dt \\ ={\lambda }_{3}t \end{gathered}$$

That is,$e^I=e^{{\lambda }_{3}t}$

By equation (3.9), we can write,

$$\begin{gathered} N_3{e}^I=\int \left[\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)\right]e^{{\lambda }_{3}t}dt \\ =\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\int \left[e^{\left({\lambda }_3-{\lambda }_1\right)t}-e^{\left({\lambda }_3-{\lambda }_2\right)t}\right]dt \\ =\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\left[\frac{e^{\left({\lambda }_3-{\lambda }_1\right)t}}{{\lambda }_3-{\lambda }_1}-\frac{e^{\left({\lambda }_3-{\lambda }_2\right)t}}{{\lambda }_3-{\lambda }_2}\right]+C \end{gathered}$$

Hence,

$N_3=\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\left[\frac{e^{-{\lambda }_{1}t}}{{\lambda }_3-{\lambda }_1}-\frac{e^{-{\lambda }_{2}t}}{{\lambda }_3-{\lambda }_2}\right]+C{e}^{-{\lambda }_{3}t}$ …… (1)

Find the value of C, using the initial condition.

That is, take $t=0,N_3=0$, to get,

$$\begin{gathered} 0=\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\left[\frac{1}{{\lambda }_3-{\lambda }_1}-\frac{1}{{\lambda }_3-{\lambda }_2}\right]+C \\ C=-\frac{{\lambda }_1{\lambda }_2{N}_0}{{\lambda }_2-{\lambda }_1}\left[\frac{1}{{\lambda }_3-{\lambda }_2}-\frac{1}{{\lambda }_3-{\lambda }_1}\right] \\ =\frac{{\lambda }_1{\lambda }_2{N}_0}{\left({\lambda }_3-{\lambda }_1\right)\left({\lambda }_3-{\lambda }_2\right)} \end{gathered}$$

Substitute back in equation (1), to get,

$N_3={\lambda }_1{\lambda }_2{N}_0\left[\frac{e^{-{\lambda }_{1}t}}{\left({\lambda }_2-{\lambda }_1\right)\left({\lambda }_3-{\lambda }_1\right)}+\frac{e^{-{\lambda }_{2}t}}{\left({\lambda }_1-{\lambda }_2\right)\left({\lambda }_3-{\lambda }_2\right)}+\frac{e^{-{\lambda }_{3}t}}{\left({\lambda }_1-{\lambda }_3\right)\left({\lambda }_2-{\lambda }_3\right)}\right]$