Question

In Problem 1 suppose that time is measured in days, that the decay constants are ${\mathbf{k}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{138629}$and ${\mathbf{k}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{004951}$, and that localid="1663912510521" ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{20}$. Use a graphing utility to obtain the graphs of the solutions x(t), y(t) and z(t) on the same set of coordinate axes. Use the graphs to approximate the half-lives of substances X and Y.

Short answer

Therefore, the half-life of substance Y is approximately 172 days. And the graphs are shown as below:

Step-by-step solution

Some equations

Let x(t), y(t) and z(t) be the amounts of substances X, Y and Z at time T. And ${\mathit{k}}_{\mathbf{1}}\mathbf{=}\mathbf{-}{\mathit{\lambda }}_{\mathbf{1}}\mathbf{<}\mathbf{0}\mathbf{,}{\mathit{k}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{<}\mathbf{0}$are the decay constants for substances X, Y.

A model of the radioactive decay series for three elements is the linear system of three first order differential equations which are shown below:

role="math" localid="1663908614897" $$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}{\mathit{\lambda }}_{\mathbf{1}}\mathit{x}\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{\lambda }}_{\mathbf{1}}\mathit{x}\mathbf{-}{\mathit{\lambda }}_{\mathbf{2}}\mathit{y}\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{\lambda }}_{\mathbf{2}}\mathit{y}\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

Now solve the linear system using the initial conditions $\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{x}}_{\mathbf{0}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Integrate the above equation

The equation (1) can be solved by using separation of variables as shown below.

$\frac{dx}{x}=-{\lambda }_{1}dt$

Integrate on both the sides.

$\begin{array}{rcl}\int \frac{dx}{x}& =& -{\lambda }_1\int dt\\ \ln \left(x\right)& =& -{\lambda }_{1}t+c\\ x& =& e^{-{\lambda }_{1}t+c}\\ x& =& e^{-{\lambda }_{1}t}·e^c\\ x& =& c_1{e}^{-{\lambda }_{1}t}\left[\text{Take}{e}^c=c_1,\text{a constant.}\right]\end{array}$

Use initial condition

Use the initial condition $x\left(0\right)=x_0$and solve for the constant $c_1$.

$$\begin{gathered} x=c_1{e}^{-{\lambda }_{1}t} \\ x\left(t\right)=c_1{e}^{-{\lambda }_{1}t} \\ x\left(0\right)=c_1{e}^{-{\lambda }_1\left(0\right)} \\ {x}_0=c_1{e}^{-0} \end{gathered}$$

Since $\left.\left[x\right(0)=x_0\right]$

$x_0=c_1$

Thus, $x\left(t\right)=x_0{e}^{-{\lambda }_{1}t}$

Solve for y

Solve the equation (2) for y.

Substitute $x\left(t\right)=x_0{e}^{-{\lambda }_{1}t}$in the equation (2).

$$\begin{gathered} \frac{dy}{dt}={\lambda }_{1}x-{\lambda }_{2}y \\ \frac{dy}{dt}={\lambda }_1{x}_0{e}^{-{\lambda }_{1}t}-{\lambda }_{2}y \\ \frac{dy}{dt}+{\lambda }_{2}y={\lambda }_1{x}_0{e}^{-{\lambda }_{1}t}\dots \dots \left(4\right) \end{gathered}$$

Compare the equation (4) with a linear equation $\frac{dy}{dt}+P\left(t\right)y=Q\left(t\right)$.

Find the integrating factor

$\begin{array}{rcl}{e}^{\int P\left(t\right)dt}& =& e^{\int {\lambda }_{2}dt}\\ & =& e^{{\lambda }_{2}t}\end{array}$

Multiply (4) with the integrating factor.

$$\begin{gathered} e^{{\lambda }_{2}t}\frac{dy}{dt}+{\lambda }_2{e}^{{\lambda }_{2}t}y={\lambda }_{1}x \\ \frac{d}{dt}\left[e^{{\lambda }_{2}t}y\right]={\lambda }_1{x}_0{e}^{\left({\lambda }_2-{\lambda }_1\right)t} \end{gathered}$$

Integrate with respect to t on both sides.

$\begin{array}{rcl}\int \frac{d}{dt}\left[e^{{\lambda }_{2}t}y\right]dt& =& \int \left[{\lambda }_1{x}_0{e}^{\left({\lambda }_2-{\lambda }_1\right)t}\right]\\ dt{e}^{{\lambda }_{2}t}y& =& \left({\lambda }_1{x}_0\right)\left(\frac{e^{\left({\lambda }_2-{\lambda }_1\right)t}}{{\lambda }_2-{\lambda }_1}\right)+c_2\\ e^{{\lambda }_{2}t}y& =& \left({\lambda }_1{x}_0\right)\left(\frac{e^{{\lambda }_{2}t}·e^{-{\lambda }_{1}t}}{{\lambda }_2-{\lambda }_1}\right)+c_2\\ y& =& \frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}\right)+c_2{e}^{-{\lambda }_{2}t}\end{array}$

Solve for constant

Use the initial condition $y\left(0\right)=0$and solve for the constant $c_2$.

$\begin{array}{rcl}y\left(t\right)& =& \frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}\right)+c_2{e}^{-{\lambda }_{2}t}\\ y\left(0\right)& =& \frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_1\left(0\right)}+c_2{e}^{-{\lambda }_2\left(0\right)}\\ 0& =& \frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-0}+c_2{e}^{-0}\\ & =& \frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}+c_2\\ c_2& =& \frac{-{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}\end{array}$

Thus,$y\left(t\right)=\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)$

Solve the equation (3) for z.

Substitute $y\left(t\right)=\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)$ in the equation (3).

$$\begin{gathered} \frac{dz}{dt}={\lambda }_{2}y \\ \frac{dz}{dt}=\frac{{\lambda }_1{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right) \end{gathered}$$

Use variable separable method.

$dz=\frac{{\lambda }_1{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)dt$

Integrate on both sides.

$$\begin{gathered} \int dz=\int \frac{{\lambda }_1{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)dt \\ z=\frac{{\lambda }_1{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}\int \left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)dt \\ =\frac{{\lambda }_1{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}\left(\frac{e^{-{\lambda }_{1}t}}{-{\lambda }_1}-\frac{e^{-{\lambda }_{2}t}}{-{\lambda }_2}\right)+c_3 \\ z\left(t\right)=-\frac{{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{1}t}+\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{2}t}+c_3 \end{gathered}$$

Use initial condition

Use the initial condition $z\left(0\right)=0$ and solve for the constant $c_3$.

$\begin{array}{rcl}z\left(t\right)& =& -\frac{{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{1}t}+\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{2}t}+c_3\\ z\left(0\right)& =& \frac{-{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_1\left(0\right)}+\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_2\left(0\right)}+c_3\\ & =& \frac{-{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}+\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}+c_3\\ c_3& =& \frac{x_0\left({\lambda }_2-{\lambda }_1\right)}{{\lambda }_2-{\lambda }_1}\\ & =& x_0\end{array}$

Thus,

$$\begin{gathered} z=\frac{-{\lambda }_2{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{1}t}+\frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{2}t}+x_0 \\ =x_0\left[1-\frac{{\lambda }_2}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{1}t}+\frac{{\lambda }_1}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{2}t}\right] \end{gathered}$$

Therefore, the required solution of given system is

$\begin{array}{rcl}x\left(t\right)& =& x_0{e}^{-{\lambda }_{1}t}\\ y\left(t\right)& =& \frac{{\lambda }_1{x}_0}{{\lambda }_2-{\lambda }_1}\left(e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}\right)\\ z\left(t\right)& =& x_0\left[1-\frac{{\lambda }_2}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{1}t}+\frac{{\lambda }_1}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_{2}t}\right]\end{array}$

Find value of solution

Given that the time is measured in days and the decay constants are $k_1=-0.138629$ and $k_2=-0.004951.$

Since $k_1=-{\lambda }_1<0,k_2=-{\lambda }_2<0$, that implies ${\lambda }_1=0.138629$and ${\lambda }_2=0.004951$.

Also given that $x_0=20$

Now substitute all the known values in solution of the linear system.

$$\begin{gathered} x\left(t\right)=20e^{-(0.138629)t} \\ y\left(t\right)=\frac{20(0.138629)}{0.004951-0.138629}\left(e^{-(0.138629)t}-e^{-(0.004951)t}\right) \\ z\left(t\right)=20\left[1-\frac{0.004951}{0.004951-0.138629}{e}^{-(0.138629)t}+\frac{0.138629}{0.004951-0.138629}{e}^{-(0.004951)t}\right] \end{gathered}$$

Draw the graph

Use a graphing utility to draw the solutions x(t), y(t) and z(t) on the same coordinate axes.

$$\begin{gathered} x\left(t\right)=20e^{-(0.138629)t} \\ y\left(t\right)=-20.74073520\left(e^{-(0.138629)t}-e^{-(0.004951)t}\right) \\ z\left(t\right)=20+0.7407351996e^{-(0.138629)t}-20.74073520e^{-(0.004951)t} \end{gathered}$$

Here, $x\left(t\right)⩾0,y\left(t\right)⩾0$and $z\left(t\right)⩾0$since x, y, z are amounts of the elements X, Y, Z which are radioactive substances.

The graphs of x(t), y(t) and z(t) are shown below:

Approximate the half-lives of substances X and Y using the above graphs.

Observe the below figure:

From the graph observe that the maximum amount x(t) of the substance X is 20 and become reduced to half of the amount 10 when t = 5.

Therefore, the half-life of substance X is approximately 5 days.

From the graph observe that the maximum amount y(t) of the substance Y is 17 and become reduced to half of the amount 8.5 when t = 172.