Question

1. Consider the following initial value problem 3t2-y, y(0) = 1 Using Euler's Method and the second order Runge-Kutta method, for t E [0, 1] with a step size of h 0.05, approximate the solution to the initial value problem. Plot the true solution and the approximate solutions on the same figure. Be sure to label your axis and include an a. appropriate legend b. Verify that the analytic solution to the differential equation is given by y(t) 3t2 -6t - 5e+6. c. For each approximation method, calculate the local truncation error (LTE) at each time step. Plot the LTE for each method on the same more accurate approximation? figure. Which technique yields the

Complete using MatLab

Answer 1

%%%%%%%%%%%% Solution of (a) and (b) by Euler

clc;
clear all;

f=@(t,y) 3*t^2-y;
a=0;% starting point
b=1; %ending point
h=0.05; % step length

n=(b-a)/h;
y(1)=1; % Initial condition
t=a:h:b;

%Euler formula
for i=1:n
y(i+1)=y(i)+h*f(t(i),y(i));
end
disp('Approximation solution')
for i=1:n+1
fprintf('y(%f)= \t %10f \n',t(i),y(i))
end
%True solution
for i=1:n+1
y_ex(i)=3*t(i)^2-6*t(i)-5*exp(-t(i))+6;
end
  
plot(t,y,'-r*')
hold on
plot(t,y_ex,'-m')

xlabel('t','fontsize',15)

ylabel('y','fontsize',15)
legend('Euler method','True solution')

disp('________________________________________________________________________________')
disp('t Euler solution True solution ')
disp('________________________________________________________________________________')

for i=1:n+1
fprintf('%f \t %10f \t %15f \n',t(i),y(i), y_ex(i))
end

Approximation solution
y(0.000000)=    1.000000
y(0.050000)=    0.950000
y(0.100000)=    0.902875
y(0.150000)=    0.859231
y(0.200000)=    0.819645
y(0.250000)=    0.784662
y(0.300000)=    0.754804
y(0.350000)=    0.730564
y(0.400000)=    0.712411
y(0.450000)=    0.700790
y(0.500000)=    0.696126
y(0.550000)=    0.698820
y(0.600000)=    0.709254
y(0.650000)=    0.727791
y(0.700000)=    0.754776
y(0.750000)=    0.790538
y(0.800000)=    0.835386
y(0.850000)=    0.889616
y(0.900000)=    0.953511
y(0.950000)=    1.027335
y(1.000000)=    1.111343
________________________________________________________________________________
t Euler solution True solution
________________________________________________________________________________
0.000000    1.000000    1.000000
0.050000    0.950000    0.951353
0.100000    0.902875    0.905813
0.150000    0.859231    0.863960
0.200000    0.819645    0.826346
0.250000    0.784662    0.793496
0.300000    0.754804    0.765909
0.350000    0.730564    0.744060
0.400000    0.712411    0.728400
0.450000    0.700790    0.719359
0.500000    0.696126    0.717347
0.550000    0.698820    0.722751
0.600000    0.709254    0.735942
0.650000    0.727791    0.757271
0.700000    0.754776    0.787073
0.750000    0.790538    0.825667
0.800000    0.835386    0.873355
0.850000    0.889616    0.930425
0.900000    0.953511    0.997152
0.950000    1.027335    1.073795
1.000000    1.111343    1.160603
>>

* - Euler method True solution 0.6 L 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0. 9 1

%%%%%%%%%%%% Solution of (a) and (b) by Rk2

clc;
clear all;

f=@(t,y) 3*t^2-y;
a=0;% starting point
b=1; %ending point
h=0.05; % step length
n=(b-a)/h;
y(1)=1; % Initial condition
t=a:h:b;

for i=1:n
m1=h*f(t(i),y(i));
m2=h*f(t(i+1),y(i)+m1);
y(i+1)=y(i)+(m1+m2)/2;
end

disp('Approximation solution by Rk2')
for i=1:n+1
fprintf('y(%f)= \t %10f \n',t(i),y(i))
end
%True solution
for i=1:n+1
y_ex(i)=3*t(i)^2-6*t(i)-5*exp(-t(i))+6;
end
  
plot(t,y,'-r*')
hold on
plot(t,y_ex,'-m')

xlabel('t','fontsize',15)
ylabel('y','fontsize',15)
legend('RK2 method','True solution')

disp('________________________________________________________________________________')
disp('t RK2 solution True solution ')
disp('________________________________________________________________________________')

for i=1:n+1
fprintf('%f \t %10f \t %15f \n',t(i),y(i), y_ex(i))
end

Approximation solution by Rk2
y(0.000000)=    1.000000
y(0.050000)=    0.951438
y(0.100000)=    0.905983
y(0.150000)=    0.864216
y(0.200000)=    0.826689
y(0.250000)=    0.793925
y(0.300000)=    0.766425
y(0.350000)=    0.744661
y(0.400000)=    0.729087
y(0.450000)=    0.720132
y(0.500000)=    0.718204
y(0.550000)=    0.723691
y(0.600000)=    0.736964
y(0.650000)=    0.758375
y(0.700000)=    0.788257
y(0.750000)=    0.826930
y(0.800000)=    0.874695
y(0.850000)=    0.931841
y(0.900000)=    0.998642
y(0.950000)=    1.075358
y(1.000000)=    1.162238
________________________________________________________________________________
t RK2 solution True solution
________________________________________________________________________________
0.000000    1.000000    1.000000
0.050000    0.951438    0.951353
0.100000    0.905983    0.905813
0.150000    0.864216    0.863960
0.200000    0.826689    0.826346
0.250000    0.793925    0.793496
0.300000    0.766425    0.765909
0.350000    0.744661    0.744060
0.400000    0.729087    0.728400
0.450000    0.720132    0.719359
0.500000    0.718204    0.717347
0.550000    0.723691    0.722751
0.600000    0.736964    0.735942
0.650000    0.758375    0.757271
0.700000    0.788257    0.787073
0.750000    0.826930    0.825667
0.800000    0.874695    0.873355
0.850000    0.931841    0.930425
0.900000    0.998642    0.997152
0.950000    1.075358    1.073795
1.000000    1.162238    1.160603
>>

1.15 RK2 method -True solution 1.05 >0.95 0.9 0.8 0.7 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0. 9 1

C)

clc;
clear all;

f=@(t,y) 3*t^2-y;
a=0;% starting point
b=1; %ending point
h=0.05; % step length
n=(b-a)/h;
y(1)=1; % Initial condition
t=a:h:b;

%Euler formula
for i=1:n
y(i+1)=y(i)+h*f(t(i),y(i));
end
y1(1)=1
for i=1:n
m1=h*f(t(i),y1(i));
m2=h*f(t(i+1),y1(i)+m1);
y1(i+1)=y1(i)+(m1+m2)/2;
end

%True solution
for i=1:n+1
y_ex(i)=3*t(i)^2-6*t(i)-5*exp(-t(i))+6;
end
  
for i=1:n+1
L(i)= abs(y(i)-y_ex(i));
end
  
for i=1:n+1
L1(i)= abs(y1(i)-y_ex(i));
end
  
plot(t,L,'-r*')
hold on
plot(t,L1,'-m')

xlabel('t','fontsize',15)
ylabel('LTE','fontsize',15)
legend('Euler','Rk2')

0.05 0.045 -Euler -Rk2 0.04 0.035 0.03 E 0.025 0.02 0.015 0.01 0.005 OL 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0. 9 1