Question

Solve the DE for x(t) given the following DE and volume solution of V(t)

then answer the case1 and case 2 questions

V(t)=180-100e^-0.01t+20e^-0.05t

^{dx/dt = i(t) – r(t) (1)}

Case 1 Let i(t) = e^-0.01t and r(t) = e^-0.05t

             Solve for x(t) and plot a graph for x(t) and the function V(t)

            What is the limiting value of x(t) that is what is x(t) as t goes to infinity.

             How does the solution vary as a function given the initial conditions of X₀=0, 80, 200, 400, 500

            What initial salt content X₀ would you select if you want a salt content output of 350 to be reached at t=50

Case 2 Let i(t) =0.2 and r(t) = (1/5)/(1+t²)

             Solve for x(t) and plot a graph for x(t) and the function V(t)

             Is there a limiting value for x(t)?

             How does the solution vary as a function given the initial conditions of X₀=0, 80, 200, 400, 500

             What initial condition X₀ yields a minimum concentration that is less than 50 at time t=100

Answer 1

for Case 1

S X - Search Documentation GO E É CE X TILL ILL MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW D E L Find Files Insert & fx Fri > PL Run Section 3 Compare - Go To Comment % 99% New Open Save Breakpoints Run Run and Advance Run and Print Find Indent - - Advance Time FILE NAVIGATE EDIT BREAKPOINTS RUN ha L C Program Files ▸ MATLAB MATLAB Production Server R2015a bin Editor - C:\Program Files\MATLAB\MATLAB Production Server\R2015a\bin\or124578.m odetry1. mx case1_lin. mxor124578. mx + clc; close all; clear all; tspan=[0:50:1000]; $declares the time interval n=1000/50; xo=[0 80 200 400 500); different initial values [t, zl]=ode45 (@(t, x) casel_lin(t,x), tspan, xo (1)); } solves the differential equation for xo=0 pl-plot(t, zl, 'green'); } plots the values of x vs t for xo=0 hold on; grid on; [t, 22]=ode 45 (@(t, x) casel_lin(t, x),tspan, xo (2)); } solves the differential equation for xo=80 p2=plot(t, 22, 'blue'); $ plots the values of x vs t for xo=80 hold on; [t, 23]=ode 45 (@(t, x) casel_lin(t, x), tspan, xo (3)); } solves the differential equation for xo=200 pl-plot(t, 23, 'red'); } plots the values of x vs t for xo=200 hold on; [t,24]=ode 45 (@(t, x) casel_lin(t,x), tspan, xo (4)); } solves the differential equation for xo=400 p4=plot(t, 24, 'black'); plots the values of x vs t for xo=400 hold on; [t,z5]=ode 45 (@ (t, x) casel_lin(t,x), tspan, xo (5));% solves the differential equation for xo=500 Q for j=1:n+1 V(3)= (180-(100* (exp(-.01*t (5))))+ (20* (exp(-0.05*t (1))))); calculates the value of V(t) end p5=plot(t, z5,'yellow'); plots the values of x vs t for xo=500 hold on; p6=plot(t,V); % plots the values of V vst legend('x0=0','x0=80','x=200','x0=400','x0=500","V (t)'); L LLL LLL LLL script In 26 Col 43 21:49 A 1) E - ENG, 23-03-20202

S X - Search Documentation GO E É CE PL Run and Advance Run Section Advance Run and Time RUN X MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW D E L Find Files Insert & fx Fri > 3 Compare - Go To Comment % 99% New Open Save Breakpoints Run Print Find Indent - - FILE NAVIGATE EDIT BREAKPOINTS ha L C Program Files ▸ MATLAB MATLAB Production Server R2015a bin Editor - C:\Program Files\MATLAB\MATLAB Production Server\R2015a\bin\or124578.m odetry1. mx case1_lin.m xor124578. m x + title('x v/s t'); xlabel('time'), ylabel('x'); for i=1:1000 if t(i)>=500 zl=zl(i); 22=22 (i); (i); 4 (i); 25=z5 (i); break; end end N N w N N IIIIIIIIIIIIIIIIIIIIIIIII fprintf('the value of x at t tends to infinity for initial value of x as $d is $f\n',xo (1),z1); fprintf('the value of x at t tends to infinity for initial value of x as $d is $f\n',xo (2), 22); fprintf("the value of x at t tends to infinity for initial value of x as $d is f\n',xo (3),23); fprintf('the value of x at t tends to infinity for initial value of x as $d is $f\n',xo (4), 24); fprintf('the value of x at t tends to infinity for initial value of x as $d is f\n',xo (5), Z5); for i=1:1000 if t(i)==50 zl=zl(i); 22 (i); 3(i); Z4=z4 (i); 25=z5 (i); break; end NNN N script Ln 26 col 43 21:50 23-03-20202 A 1) ENG

end fprintf('\n'); fprintf('the value of x at t =50 for initial value of x as $d is $f\n',xo (1), 21); fprintf('the value of x at t =50 for initial value of x as $d is $f\n',xo (2),z2); fprintf('the value of x at t =50 for initial value of x as $d is f\n', xo(3),23); fprintf('the value of x at t =50 for initial value of x as $d is $f\n',xo (4), 24); fprintf('the value of x at t =50 for initial value of x as $d is $f\n',xo (5), 25); script Ln 26 Col 43 21:50 ENG 23-03-2020 2 ^ 4) =

The response is

God gË 5 CE Search Docum MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW D E L Find Files Insert & fx Fri > Compare - Save Open Go To New Comment % % Breakpoints Run Print Find Indent - - FILE NAVIGATE EDIT BREAKPOINTS ha L C Program Files ▸ MATLAB MATLAB Production Server R2015a bin Command Window New to MATLAB? See resources for Getting Started. PL Run and Advance Run Section Advance Run and Time RUN the value of x at t tends to infinity for initial value of x as 0 is 96.518635 the value of x at t tends to infinity for initial value of x as 80 is 162.763569 the value of x at t tends to infinity for initial value of x as 200 is 262.130523 the value of x at t tends to infinity for initial value of x as 400 is 427.736844 the value of x at t tends to infinity for initial value of x as 500 is 510.541277 the value of x at t =50 for initial value of xas o is 37.125286 the value of x at t =50 for initial value of x as 80 is 104.217538 the value of x at t =50 for initial value of x as 200 is 204.831467 the value of x at t =50 for initial value of x as 400 is 372.533840 the value of x at t =50 for initial value of x as 500 is 456.403037

The value of xo should be 400

- O X Figure 1 File Edit View Insert Dadah Tools Desktop Window Help RA @ 09 - 0 x vist xo=0 Xo=80 xo=200 x0=400 x0=500 V(t) x 300 100 0 100 200 300 400 600 700 800 900 1000 500 time

Case 2:

S X - Search Documentation GO E É CE X TILL MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW D E L Find Files Insert & fx Fri > PL Run Section 3 Compare - Go To Comment % 99% New Open Save Breakpoints Run Run and Advance Run and Print Find Indent - - Advance Time FILE NAVIGATE EDIT BREAKPOINTS RUN ha L C Program Files ▸ MATLAB MATLAB Production Server R2015a bin Editor - C:\Program Files\MATLAB\MATLAB Production Server\R2015a\bin\or124578.m odetry1.m X case1_lin.m x or124578. m x case2_lin. m x + clc; close all; clear all; tspan=[0:50:1000]; $declares the time interval n=1000/50; xo=[0 80 200 400 500); different initial values [t, zl]=ode45 (@(t, x) case2_lin(t,x), tspan, xo (1)); } solves the differential equation for xo=0 pl-plot(t, zl, 'green'); } plots the values of x vs t for xo=0 hold on; grid on; [t, 22]=ode 45 (@(t, x) case2_lin(t, x),tspan, xo (2)); } solves the differential equation for xo=80 p2=plot(t, 22, 'blue'); $ plots the values of x vs t for xo=80 hold on; [t, 23]=ode 45 (@(t, x) case2_lin(t, x), tspan, xo (3)); } solves the differential equation for xo=200 pl-plot(t, 23, 'red'); } plots the values of x vs t for xo=200 hold on; [t,24]=ode 45 (@(t, x) case2_lin(t,x), tspan, xo (4)); } solves the differential equation for xo=400 p4=plot(t, 24, 'black'); plots the values of x vs t for xo=400 hold on; [t,z5]=ode 45 (@ (t, x) case2_lin(t,x), tspan, xo (5));% solves the differential equation for xo=500 Q for j=1:n+1 V(3)= (180-(100* (exp(-.01*t())))) + (20* (exp(-0.05*t (1))))); calculates the value of V(t) end p5=plot(t, z5,'yellow'); plots the values of x vs t for xo=500 hold on; p6=plot(t,V); plots the values of V vs t legend('x0=0','x0=80','x=200','x0=400','x0=500","V (t)'); L LLL LLL LLL script Ln 56 Col 35 22:03 - ENG, 23-03-20202 A 1)

S X - Search Documentation GO E É CE X N N w N N MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW D E L Find Files Insert & fx Fri > PL Run Section 3 Compare - Go To Comment % 99% New Open Save Breakpoints Run Run and Advance Run and Print Find Indent - - Advance Time FILE NAVIGATE EDIT BREAKPOINTS RUN ha L C Program Files ▸ MATLAB MATLAB Production Server R2015a bin Editor - C:\Program Files\MATLAB\MATLAB Production Server\R2015a\bin\or124578.m odetry1.m x case1_lin.m x or124578. m x case2_lin.m x + title('x v/s t'); xlabel('time'), ylabel('x'); for i=1:1000 if t(i)>=500 zl=zl(i); 22=22 (i); (i); 4 (i); 25=z5(i); break; end end fprintf('there is no limiting value of x'); for i=1:1000 if t(i)==100 Zl=zl(i); 22=22 (i); 23=23 (i); Z4=24 (i); 25=25 (i); break; end end fprintf('\n'); fprintf('the value of x at t =100 for initial value of x as $d is $f\n',xo (1),21); fprintf('the value of x at t =100 for initial value of x as $d is $f\n',xo (2),z2); fprintf('the value of x at t =100 for initial value of x as $d is $f\n',xo (3),23); IIIIIIIIIIIIIIIIIIIIIIIIII script Ln 56 Col 35 22:03 - ENG, 23-03-20202 A 1)

55 - fprintf('the value of x at t fprintf('the value of x at t =100 for initial value of x as $d is $f\n',xo (4), 24); =100 for initial value of x as $d is $f\n',xo (5),25);

The respone is

S X - Search Documentation GO E È CE MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW D E L Find Files Insert & fx Fri > Compare - Save Open Go To New Comment % % Breakpoints Run Print Find Indent - - FILE NAVIGATE EDIT BREAKPOINTS a o Program Files ▸ MATLAB MATLAB Production Server R2015a bin Command Window New to MATLAB? See resources for Getting Started. PL Run and Advance Run Section Advance Run and Time RUN there is no limiting value of x the value of x at t -100 for initial value of x as 0 is 19.998269 the value of x at t =100 for initial value of x as 80 is 99.251355 the value of x at t =100 for initial value of x as 200 is 218.129871 the value of x at t =100 for initial value of x as 400 is 416.260732 the value of x at t =100 for initial value of x as 500 is 515.326162 A la 4) = 22:05 ENG 23-03-20202

Thus there is no limiting value of x and the initial x should be taken here as 0 to keep x less tha 50 for t=100

- 5 X Figure 1 File Edit View 18 Insert Tools Desktop Window 3 Help E x vlst xo=0 x0=80 X0=200 x0=400 X0=500 V(t) X 100 200 300 400 600 700 800 900 1000 500 time 22:06 Alfa ) ENG 23-03-20202

The plot for case 2