(Solved): Write and submit a computer code (MATLAB is preferred) that plots the solution of the point ki ...

Write and submit a computer code (MATLAB is preferred) that plots the solution of the point kinetics equations for finite, multiplying systems with constant source term, the code should output a plot of the number of neutrons in the system as a function of time \( (n(t)) \) : a. The code input should be in excel, outputs a plot of the number of neutrons and an excel sheet of \( n(t) \), concentration of delayed neutron precursors ( \( C(t) \) ). b. Input: Source term, multiplication factor, neutron lifetime, delayed neutron groups, half-lifes, fraction..etc (ALL INFORMATION MUST BE INPUT TO THE CODE VIA AN EXCEL INPUT SHEET). Test and report: Test your code by solving problem \( 5.15 \) in the book. Submit: A report of the test results, the code, the input file (excel sheet), a short description of your code and input file structure. \( [5.15]^{*} \) Solve the kinetics equations numerically with one group of delayed neutrons, given in problem [5.10], using the data \( \beta=0.007, \Lambda=5 \cdot 10^{-5} \mathrm{~s} \). and \( \lambda=0.08 \mathrm{~s}^{-1} \) a. For a step insertion of \( +0.25 \) dollars between 0 and \( 5 \mathrm{~s} \). b. For a step insertion of \( -0.25 \) dollars between 0 and \( 5 \mathrm{~s} \). Assume that the reactor is initially in the critical state, and plot \( n(t) / n(0) \) for each case. The kinetics equations are set equal to zero to find the initial conditions. \[ \mathrm{c}:=\frac{\beta \cdot \mathrm{P}_{0}}{\lambda \cdot \Lambda}=1.75 \times 10^{3} \] Use a program such as Mathematica, Matlab, or in this case Mathcad 14 to solve the pair of differential equations. It is important to note that due to the effects of prompt neutrons, it will be best to solve the differential equations using a stiff method, in this case, the Radau function in Mathcad. Dividing the results by the initial condition standardizes the results to obtain the graph below a) \[ \begin{array}{l} \rho_{1}:=0.25 \cdot \beta=1.75 \times 10^{-3} \\ \mathrm{D}_{1}(\mathrm{t}, \mathrm{y}):=\left[\begin{array}{c} \frac{\left(\rho_{1}-\beta\right)}{\Lambda} \mathrm{y}_{0}+\lambda \cdot \mathrm{y}_{1} \\ \left(\frac{\beta}{\Lambda}\right) \cdot \mathrm{y}_{0}-\lambda \cdot \mathrm{y}_{1} \end{array}\right] \\ \rho_{2}:=-0.25 \cdot \beta=-1.75 \times 10^{-3} \\ Y:=\operatorname{Radau}\left(\frac{\mathrm{y}}{\mathrm{P}_{0}}, 0,50,10000, \mathrm{D}_{1}\right) \\ \text { YY := Radau }\left(\frac{\mathrm{y}}{\mathrm{P}_{0}}, 0,50,10000, \mathrm{D}_{2}\right) \\ n:=0,1 \ldots 10000 \\ \end{array} \] However, taking into account the reactor's period is 50 seconds, the plot will look like the one shown below.