the question is in red, i only need help with a part of the project
Project A: Corn, cats and rats 3 Project A: Corn, cats and rats On an isolated island, some begins to grow. From the beginning there are 75 plants but soon they increase in number according to dP/dt=α1P−α2P ^2. The unit of time is years, P (t) the number of plants at time t and the constants α1 = 15 and α2 = 1.4 * 10−5. If it remains isolated, the number of plants will reach a maximum value, which then? To be solved without ODE solvers the program! But on the day when the plants reach 80% of the maximum value, four rats appear. Find with Runge-Kutta method and step length 1 day (ie 1/365) which day it is. Check if Runge-Kutta's method with the step length 1 week combined with appropriate interpolation arrives at the same day. Investigate what hour Runge-Kutta's method combined with appropriate interpolation arrives with the different ones step lengths 1 day and 1 week. The rats thrive on the island and eat up a lot of the plants. If there are too many of them, there will be a shortage of plants. The interaction between number of rats and plants can be described by dP/dt = α1*P−α2*P^2−α3*P*R dR/dt=−β1*R^1.4+β2*P^0.6*R^0.8 where α3 = 0.014, β1 = 1.9, β2 = 0.083 If the island now remains isolated, the number of plants and rats will reach constant equilibrium value, which ones? (Then use your ODE reader program to check your calculation). But, exactly 0.5 years after the rats arrived, 2 cats landed. The cats eat the rats but not the plants, so that the growth equation of the plants is not affected but the equation of the rats (and of course the cats) is: dR / dt = −β1*R^1.4 + β2*P^0.6R^0.8 - β3*RK dK / dt = −γ1*K + γ2*R*K^0.5 where β3 = 1.5, γ1 = 1.9, γ2 = 0.0231 Calculate the number of the plants, rats and cats in the different scenarios (ie when the plants came, when the rats came, when the cats came, and when it was four years later. Plot the development for all the different varieties. How much do the final values for each system differ from each other? (Then use your ODE solver software to control your calculation of final values). On the first day of the fifth year, the people come there and decide to fight the rats with poison. 65% of the rats die, but also 17% of the cats. How many plants, rats and cats do you have after 4.5 Years, ie in the middle of summer the fifth year? And at the end of the year? Was the spray (of poison) worth it? I would like written solutions for this: I want a solution for the equilibrium points using precise and efficient methods (the method used here is not efficient), an error estimation for every value in the tables below. Thank you in advance. Table 1 When plants reach 80% of the maximum value (step When plants reach 80% of the maximum value (step length one day) When plants reach 80% of the maximum value (step length one week) When plants reach 80% of the maximum value (step length one day) The maximum value of plants unit day day hour hour plants amount 266 267 6398 6408 1071429 Table 2 Time When plants and rats reach constant equilibrium values When 2 cats arrived (six months after the rats) 4.5 years (six months after humans arrived) unit plants rats plants rats plants rats cats amoun t 166928 904 169500 907 600200 165 907 Here are the codes with the answers: main code: clear; clc; close all; global alpha1 alpha2 alpha3 beta1 beta2 beta3 gamma1 gamma2 global t80DStab alpha1 = 15; alpha2 = 1.4E-5; alpha3 = 0; beta1 = 0; beta2 = 0; beta3 = 0; gamma1 = 0; gamma2 = 0; tol = 1E-1; hD = 1/365; % One day time step hW = 1/52.1429; % Week time step P0 = 75; % Initial plants R0 = 0; %Initial rats C0 = 0; % Initial cats y0 = [P0, R0, C0]'; Tf = 1; tspan = [0, Tf]; [tD, yD] = RK4(@myode, tspan, y0, hD); PStab = alpha1/alpha2; idxStab = find(abs(yD(:,1) - PStab) <= tol,1);="" td(idxstab)="" %="" analytical="" solution="" for="" plant="" only="" system="" %fsol="@(t)" 1125*exp(15*t)/((exp(15*t)-1)*0.00105+15);="" fprintf('\n\nwhen="" solved="" analytically,="" the="" plants="" grow="" stable="" to="" %10.3f="" plants="" in="" %4.3f="" years(%5.1f="" days),="" in="" absense="" of="" rats.\n\n',pstab,="" td(idxstab),td(idxstab)*365);="" %="" find="" time="" at="" which="" 80%="" of="" this="" stable="" value="" is="" reached="" idx80dstab="find((yD(:,1)-0.8*PStab)">= 0,1); t80DStab = tD(idx80DStab); [tW, yW] = RK4(@myode, tspan, y0, hW); idx80WStab = find((yW(:,1)-0.8*PStab) >= 0,1); t80DWStab = (tW(idx80WStab-1)+((0.8*PStab - yW(idx80WStab-1,1))/(yW(idx80WStab,1)-yW(idx80WStab-1,1)))*(tW(idx80WStab)-tW(idx80WStab-1))); fprintf('\n\nDay interpolation from week time step is %6.2f and from day time step is %6.4f. Clearly, they dont match\n\n',t80DWStab*365,t80DStab*365); % RATS come in tspan = [t80DStab, 2]; P01 = yD(idx80DStab,1); R01 = 4; C01 = 0; y01 = [P01, R01, C01]'; alpha3 = 0.014; beta1 = 1.9; beta2 = 0.083; [tD1, yD1] = RK4(@myode, tspan, y01, hD); Y = cat(1,yD(1:idx80DStab-1,:),yD1); T = cat(1,tD(1:idx80DStab-1,:),tD1); fprintf('\n\nWhen 4 rats are introduced, the plants stabilize to %8.2f and rats stabilize to %5.2f.\n\n',Y(end,1),Y(end,2)); % And the CATS come idxp5yr = find(tD1-tD1(1)>=0.5,1); tspan = [tD1(idxp5yr) 4]; P02 = yD1(idxp5yr,1); R02 = yD1(idxp5yr,2); C02 = 2; y02 = [P02 R02 C02]'; beta3 = 1.5; gamma1 = 1.9; gamma2 = 0.0231; [tD2, yD2] = RK4(@myode, tspan, y02, hD); Y = cat(1,cat(1,yD(1:idx80DStab-1,:),yD1(1:idxp5yr-1,:)),yD2); T = cat(1,cat(1,tD(1:idx80DStab-1,:),tD1(1:idxp5yr-1)),tD2); % And the POISON is applied by evil humans tspan = [T(end)+hD 4.5]; P03 = Y(end,1); R03 = Y(end,2)*0.35; C03 = Y(end,3)*0.83; y03 = [P03 R03 C03]'; beta3 = 1.5; gamma1 = 1.9; gamma2 = 0.0231; [tD3, yD3] = RK4(@myode, tspan, y03, hD); Y = cat(1,Y,yD3); T = cat(1,T,tD3); str = 'Population dynamics plot'; figure('Numbertitle','off','Name',str); plot(T,Y(:,1),'Color',[1,0,0],'LineWidth',2,'DisplayName','Plants'); hold on; plot(T,Y(:,2),'Color',[0,1,0],'LineWidth',2,'DisplayName','Rats'); plot(T,Y(:,3),'Color',[0,0,1],'LineWidth',2,'DisplayName','Cats'); grid on; legend; xlabel('Time(years)'); ylabel('Species population'); title(str); fprintf('\n\nClearly, the rats are again rising in population, which is why, the poison application was really not effective.\n\n'); differential equations : function dy = myode(t,y) %My differential equation system global alpha1 alpha2 alpha3 beta1 beta2 beta3 gamma1 gamma2 P = y(1); % Plants R = y(2); % Rats C = y(3); % Cats dP = alpha1*P - alpha2*power(P,2) - alpha3*P*R; dR = -beta1*power(R,1.4) + beta2*power(P,0.6)*power(R,0.8) - beta3*R*C; dK = -gamma1*C + gamma2*R*power(C,0.5); dy = [dP, dR, dK]'; RK4: function [t, y] = RK4(f, tspan, y0, h) % RK4 with input step size % Inputs: % f: function handle of f(t, y) % tspan: the time period for simulation (should be a 1x2 array ... % contain start time and end time) % y0: the initial conditions for the differential equation % Outputs: % t: corresponding time sequence as a T x 1 vector % y: the solution of the differential equation as a T x n matrix, ... % where T is the number of time steps and n is the dimension of y hk = h; % Initialize error e0 = 1E-4; tk = tspan(1); y = y0; tf = tspan(2); % Compute solutions by RK4 adpative step size algorithm while tk(end) < tf hk = min(hk, tf-tk(end)); k1 = f(tk(end), y(:,end)); k2 = f(tk(end)+0.5*hk,y(:,end)+0.5*hk*k1); k3 = f(tk+0.5*hk,y(:,end)+0.5*hk*k2); k4 = f(tk(end)+hk,y(:,end)+hk*k3); ykp1 = y(:,end) + (1/6)*hk*k1 + (1/3)*hk*k2 + (1/3)*hk*k3 + (1/6)*hk*k4; tk = cat(1,tk,tk(end)+hk); y = cat(2,y,ykp1); end t = tk; y = y'; tf="" hk="min(hk," tf-tk(end));="" k1="f(tk(end)," y(:,end));="" k2="f(tk(end)+0.5*hk,y(:,end)+0.5*hk*k1);" k3="f(tk+0.5*hk,y(:,end)+0.5*hk*k2);" k4="f(tk(end)+hk,y(:,end)+hk*k3);" ykp1="y(:,end)" +="" (1/6)*hk*k1="" +="" (1/3)*hk*k2="" +="" (1/3)*hk*k3="" +="" (1/6)*hk*k4;="" tk="cat(1,tk,tk(end)+hk);" y="cat(2,y,ykp1);" end="" t="tk;" y="">