Question

A mathematical model for the area(in cm²) that a colony of bacteria (B. dendroides) occupies is given by

$\frac{dA}{dt}=A(2.128-0.0432A)$

Suppose that the initial area is.

a. Use the RK4 method with \(h=0.5\) to complete the following table

t (days)	1	2	3	4	5
A (observed)	2.78	13.53	36.30	47.50	49.40
A (approximated)

b. Use a numerical solver to graph the solution of the initial value problem. Estimate the valuesA(1), A(2), A(3), A(4) and A(5)from the graph.

c. Use separation of variables to solve the initial-value problem and compute the actual values A(1), A(2), A(3), A(4) and A(5).

$\frac{\mathbf{d}\mathbf{A}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{128}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0432}\mathit{A}\mathbf{)}$

Short answer

a. The table is shown below

b. The grap is plotted with the above table

c. $A=\frac{0.24e^{2.128t}}{1+0.004896e^{2.128t}}$is the solution of the given differential equation.

Step-by-step solution

Runga-Kutta method:

The solution of a linear differential equation of the form $y^{\text{'}}=f(x,y)$is given as follows:

$$\begin{gathered} y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\text{where} \\ {k}_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

a) Applying First and initial approximation method

$\frac{dA}{dt}=f(t,A)=A(2.128-0.0432A),A\left(0\right)=24{\text{cm}}^2$and we have to obtain the approximate values of the solution at t = 5 using RK4 method with h = 0.5 as the following:

We can obtain the following result for using the formula

$A_{n+1}=A_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$

But before that, we have to find the constant k₁ , k₂ , k₃ and k₄ at n = 0 and h = 0.5 as the following

Since we have t₀ = 0 and A₀ = 0 , then we have

$$\begin{gathered} k_1\text{=}f\left(t_0,A_0\right) \\ =A_0\left(2.128-0.0432A_0\right) \\ =0.24(2.128-0.0432\times 0.24) \\ =0.24\times 2.117632 \\ =0.5082 \end{gathered}$$

$$\begin{gathered} k_2=f\left(t_0+\frac{1}{2}h,A_0+\frac{1}{2}h{k}_1\right) \\ =\left(A_0+\frac{1}{2}h{k}_1\right)\left(2.128-0.0432\left(A_0+\frac{1}{2}h{k}_1\right)\right) \\ =\left(0.24+\frac{1}{2}\times 0.5\times 0.5082\right)\left(2.128-0.0432\left(0.24+\frac{1}{2}\times 0.5\times 0.5082\right)\right) \\ =(0.36706)(2.11214) \\ =0.7753 \end{gathered}$$

$$\begin{gathered} k_3=f\left(t_0+\frac{1}{2}h,A_0+\frac{1}{2}h{k}_2\right) \\ =\left(A_0+\frac{1}{2}h{k}_2\right)\left(2.128-0.0432\left(A_0+\frac{1}{2}h{k}_2\right)\right) \\ =\left(0.24+\frac{1}{2}\times 0.5\times 0.7753\right)\left(2.128-0.0432\left(0.24+\frac{1}{2}\times 0.5\times 0.7533\right)\right) \\ =(0.43382)(2.10926) \\ =0.9150 \end{gathered}$$

$$\begin{gathered} k_4=f\left(t_0+h,A_0+h{k}_3\right) \\ =\left(A_0+h{k}_3\right)\left(2.128-0.0432\left(A_0+h{k}_2\right)\right) \\ =(0.24+0.5\times 0.915)(2.128-0.0432(0.24+0.5\times 0.915\left)\right) \\ =(0.69752)(2.09787) \\ =1.4633 \end{gathered}$$

After that when n = 0, since we have A₀ = 0.24 and h = 0.5, then by substituting with the four constants we can obtain from equation (1) as

$$\begin{gathered} A_1=A_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =0.24+\frac{0.5}{6}[0.5082+2(0.7753)+2(0.915)+1.4633] \\ =0.24+\frac{0.5}{6}\times (5.3521) \\ =0.686 \end{gathered}$$A

After that, we can obtain the value of A for each value of t with another four constants in the table shown below until t = 5, we can have A₅ = 48.9965

After that, we can obtain the value of for each value of with another four constants until we can complete the given table as shown below.

 b) Using a numerical solver, we can graph the solution of the given differential equation and specify the value of  from  day to  days as the following

c) We have the separable differential equation

$\frac{dA}{dt}=A(2.128-0.0432A)$and we have to solve it as the following:

$$\begin{gathered} \int \frac{dA}{A(2.128-0.0432A)}=\int dt\int \left(\frac{k_1}{A}+\frac{k_2}{2.128-0.0432A}\right)dA=\int dt\int \frac{2.128k_1-0.0432k_{1}A+k_{2}A}{A(2.128-0.0432A)}dA \\ =\int dt\int \frac{2.128k_1+\left(k_2-0.0432k_1\right)A}{A(2.128-0.0432A)}dA \\ =\int dt \end{gathered}$$

Then we have

$$\begin{gathered} 2.128k_1=1,\text{then}:k_1=0.47 \\ {k}_2-0.0432k_1=0,\text{then}:k_2=0.02 \end{gathered}$$

Then we have

$$\begin{gathered} \int \left(\frac{0.47}{A}+\frac{0.02}{2.128-0.0432A}\right)dA \\ =\int dt\int \left(0.47\frac{1}{A}-0.47\frac{0.0432}{2.128-0.0432A}\right)dA \\ =\int dt0.47\ln \left(A\right)-0.47\ln (2.128-0.0432A) \\ =t+c_1\ln \left(\frac{A}{2.128-0.0432A}\right) \\ =2.128t+c_2\frac{A}{2.128-0.0432A}=c{e}^{2.128t}----\left(2\right) \end{gathered}$$

After that, to find the value of constant (c), we have to apply the initial condition (t,4) = (0,0.24) into equation (2) , then we have

$$\begin{gathered} \frac{0.24}{2.128-0.0432\times 0.24}=c{e}^0 \\ c=0.1133 \end{gathered}$$

After that, substitute with the value of constantinto equation (2), then we have

$$\begin{gathered} \frac{A}{2.128-0.0432A}=0.1133e^{2.128t} \\ A=0.24e^{2.128t}-0.004896e^{2.128t} \\ AA\left(1+0.004896e^{2.128t}\right)=0.24e^{2.128t} \end{gathered}$$

Then we have

$A=\frac{0.24e^{2.128t}}{1+0.004896e^{2.128t}}$-------(3)

is the solution of the given differential equation.

After that, we have to obtain the values A(1), A(2), A(3), A(4) and A(5) from equation (3) as the following:

At t = 1 day, we can have

$$\begin{gathered} A_1=\frac{0.24e^{2.128}}{1+0.004896e^{2.128}} \\ =1.936 \end{gathered}$$

At t = 2 days, we can have

$$\begin{gathered} A_2=\frac{0.24e^{4.256}}{1+0.004896e^{4.256}} \\ =112.582 \end{gathered}$$

At t = 3 days, we can have

$$\begin{gathered} A_3=\frac{0.24e^{6.384}}{1+0.004896e^{6.384}} \\ =36.45 \end{gathered}$$

At t = 4 days, we can have

.$$\begin{gathered} A_4=\frac{0.24e^{8.872}}{1+0.004896e^{8.872}} \\ =47.654 \end{gathered}$$

At t = 5 days, we can have

$$\begin{gathered} A_4=\frac{0.24e^{11.09}}{1+0.004896e^{11.09}} \\ =48.867 \end{gathered}$$