Question

Mixtures Solely on the basis of the physical description of the mixture problem on page 108 and in Figure 3.3.1, discuss the nature of the functions $x_1\left(t\right)$and ${\mathit{x}}_2\left(\mathit{t}\right).$What is the behavior of each function over a long period of time? Sketch possible graphs of $x_1\left(t\right)$and ${\mathit{x}}_2\left(\mathit{t}\right).$Check you conjectures by using a numerical solver to obtain numerical solution curves of (3) subject to the initial conditions${\mathit{x}}_1\left(0\right)=25,{\mathit{x}}_2\left(0\right)=0$ .

Short answer

The solution is

$$\begin{gathered} x_1\left(t\right)=\frac{25}{2}{e}^{-\frac{1}{25}t}+\frac{25}{2}{e}^{-\frac{3}{25}t} \\ {x}_2\left(t\right)=25e^{-\frac{1}{25}t}+25e^{-\frac{3}{25}t} \end{gathered}$$

Step-by-step solution

Given information

Initial conditions

$x_1\left(0\right)=25,x_2\left(0\right)=0.$

Find second order differential equation

The two differential equations describe two tanks A and B with concentrations of$x_1\left(t\right)$ and$x_2\left(t\right)$ respectively.

$$\begin{gathered} \frac{d{x}_1}{dt}=\frac{2}{25}{x}_1+\frac{1}{50}{x}_2...\left(1\right) \\ \frac{d{x}_2}{dt}=\frac{2}{25}{x}_1-\frac{2}{25}{x}_2...\left(2\right) \\ \left(\frac{2}{25}+\frac{d}{dt}\right)x_1=\frac{1}{50}{x}_2...\left(3\right) \\ \frac{2}{25}{x}_1=\left(\frac{2}{25}+\frac{d}{dt}\right)x_2...\left(4\right) \end{gathered}$$

and we must solve them using the following strategy for $x_1$and $x_2$:

Multiply the first equation shown in (3) by$\left(\frac{2}{25}+\frac{d}{dt}\right)$ and the second equation shown in (4) by data-custom-editor="chemistry" $-\frac{1}{50}$, then we have

$$\begin{gathered} {\left(\frac{2}{25}+\frac{d}{dt}\right)}^2{x}_1=\frac{1}{50}\left(\frac{2}{25}+\frac{d}{dt}\right)x_2...\left(5\right) \\ -\frac{1}{625}{x}_1=-\frac{1}{50}\left(\frac{2}{25}+\frac{d}{dt}\right)x_2...\left(6\right) \end{gathered}$$

Add equation (5) to equation (6)

$$\begin{gathered} {\left(\frac{2}{25}+\frac{d}{dt}\right)}^2{x}_1-\frac{1}{625}{x}_1=\frac{1}{50}\left(\frac{2}{25}+\frac{d}{dt}\right)x_2-\frac{1}{50}\left(\frac{2}{25}+\frac{d}{dt}\right)x_2 \\ {\left(\frac{2}{25}+\frac{d}{dt}\right)}^2{x}_1-\frac{1}{625}{x}_1=0 \\ \frac{d^2{x}_1}{d{t}^2}+\frac{4}{25}\frac{d{x}_1}{dt}+\frac{4}{625}{x}_1-\frac{1}{625}{x}_1=0 \\ \frac{d^2{x}_1}{d{t}^2}+\frac{4}{25}\frac{d{x}_1}{dt}-\frac{3}{625}{x}_1=0 \end{gathered}$$

Find general solution

To solve the differential equation (7), we must first assume that $x_1=e^{mt}$, and then differentiate with respect to t as follows:

$$\begin{gathered} \frac{d{x}_1}{dt}=m{e}^{mt}....\left(a\right) \\ \frac{d^2{x}_1}{d{t}^2}=m^2{e}^{mt}....\left(b\right) \end{gathered}$$

Substitute with equations (a) and (b) into equation (7)

$$\begin{gathered} \left(m^2+\frac{4}{25}m+\frac{3}{625}\right)e^{mt}=0 \\ {m}^2+\frac{4}{25}m+\frac{3}{625}=0 \end{gathered}$$

$e^{mt}$cannot equal to 0 , then we have the auxiliary equation as

$$\begin{gathered} m^2+\frac{4}{25}m+\frac{3}{625}=0 \\ {m}_{\mathrm{1,2}}=\frac{-\frac{4}{25}\pm \sqrt{\frac{16}{625}-\frac{12}{625}}}{2} \\ =\frac{-\frac{4}{25}\pm \sqrt{\frac{4}{625}}}{2} \end{gathered}$$

$$\begin{gathered} =\frac{-\frac{4}{25}\pm \frac{2}{25}}{2} \\ =-\frac{2}{25}\pm \frac{1}{25} \end{gathered}$$

The general solution for concentration $x_1$

$$\begin{gathered} x\left(t\right)=c_1{e}^{m_{1}t}+c_2{e}^{m_{2}t} \\ =c_1{e}^{-\frac{1}{25}t}+c_2{e}^{-\frac{3}{25}t} \end{gathered}$$

Find solution for concentration

After that, we must insert the answer for x given in equation (8) into equation (1) to discover the solution for concentration $x_1\left(t\right)$ .

$$\begin{gathered} \frac{d}{dt}\left(c_1{e}^{-\frac{1}{25}t}+c_2{e}^{-\frac{3}{25}t}\right)=-\frac{2}{25}\left(c_1{e}^{-\frac{1}{25}t}+c_2{e}^{-\frac{3}{25}t}\right)+\frac{1}{50}{x}_2 \\ -\frac{1}{25}{c}_1{e}^{-\frac{1}{25}t}-\frac{3}{25}{c}_2{e}^{-\frac{3}{25}t}=-\frac{2}{25}{c}_1{e}^{-\frac{1}{25}t}-\frac{2}{25}{c}_2{e}^{-\frac{3}{25}t}+\frac{1}{50}{x}_2 \\ -\frac{1}{25}{c}_1{e}^{-\frac{1}{25}t}-\frac{3}{25}{c}_2{e}^{-\frac{3}{25}t}+\frac{2}{25}{c}_1{e}^{-\frac{1}{25}t}+\frac{2}{25}{c}_2{e}^{-\frac{3}{25}t}=\frac{1}{50}{x}_2 \\ \frac{1}{25}{c}_1{e}^{-\frac{1}{25}t}-\frac{1}{25}{c}_2{e}^{-\frac{3}{25}t}=\frac{1}{50}{x}_2 \\ {x}_2\left(t\right)=2c_1{e}^{-\frac{1}{25}t}-2c_2{e}^{-\frac{3}{25}t}.....\left(9\right) \end{gathered}$$

Is the solution of concentration $x_2$.

The given initial conditions into equations (8) and (9)

Apply the initial condition$x_1\left(0\right)=25$into (8)

$$\begin{gathered} 25=c{{}_{1}e}^0+c{{}_{2}e}^0 \\ {c}_1+c_2=25....\left(c\right) \end{gathered}$$

Apply the initial condition$x_1\left(0\right)=25$into (8)

$$\begin{gathered} 0=2c{{}_{1}e}^0-2c{{}_{2}e}^0 \\ {c}_1-c_2=0....\left(d\right) \end{gathered}$$

Substitute constant values

To solve the equation (c) and (d)

$c_1=\frac{25}{2}$and $c_2=\frac{25}{2}$

Substitute with the value of constants $C_1$and $C_2$into equation (8) and (9)

$$\begin{gathered} x_1\left(t\right)=\frac{25}{2}{e}^{-\frac{1}{25}t}+\frac{25}{2}{e}^{-\frac{3}{25}t} \\ {x}_2\left(t\right)=25e^{-\frac{1}{25}t}-25e^{-\frac{3}{25}t} \end{gathered}$$

Conclusion

The solution is

$$\begin{gathered} x_1\left(t\right)=\frac{25}{2}{e}^{-\frac{1}{25}t}+\frac{25}{2}{e}^{-\frac{3}{25}t} \\ {x}_2\left(t\right)=25e^{-\frac{1}{25}t}+25e^{-\frac{3}{25}t} \end{gathered}$$