Matlab code for ODE problems
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I have the function (dy/dt)=-t*e^y, y(0)=0 and have to create a code using four seperate methods (Euler, Huen, Modified Euler, and RK4) in order to graph the four different method's y values from time t (-5,5) and 1 second intervals.
I created a code with each method however I cannot get them to work even individually let alone all together.
Any help would be appreciated.
clear all
close all
clc
k=1;
t=-5:k:5
y=zeros([11 1])
y(1)=1
n=numel(y)
% Euler Method
for i=1:n-1
y(i+1)=y(i)-t*exp(y(i))*k
end
% Huen Method
for i=1:n-1
fs=-t*exp(y(i))
yp(i+1)=y(i)-fs*k
fe=-t*exp(yp(i+1))
y(i+1)=y(i)+(1/2)*(fs+fe)*k
end
% Modifier Euler Method
for i=1:n-1
y(i+1/2)=y(i)-t*exp(y(i))*(k/2)
y(i+1)=y(i)-t*exp(y(i))*k
end
% RK4
for i=1:n-1
f1=-t*exp(y(i))
f2=-(t+k/2)*exp(y(i)+.5*f1*k)
f3=-(t+k/2)*exp(y(i)+.5*f2*k)
f4=-(t+k)*exp(y(i)+f3*k)
Qk=(1/6)*(f1+2*f2+2*f3+f4)
y(i+1)=y(i)+Qk*k
end
plot(t,y);
grid on
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回答 (1 件)
Pratyush Roy
2020 年 10 月 30 日
Hi Garrett,
For all the methods for solving ODEs, the time as well as variable should change in steps. Considering the code snippet for the Euler method:
y(i+1)=y(i)-t*exp(y(i))*k
The right hand side of the equation contains multiplication of y(i) with vector t which gives us a vector but that is assigned to a scalar value y(i+1).The time values should also vary in steps with the functional value y(i). The following code snippet demonstrates the case where both time and functional value is varying:
y(i+1)=y(i)-t(i)*exp(y(i))*k
The same changes are applicable for the other techniques as well.
Hope this helps!
Regards,
Pratyush.
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