Evaluating a 2nd order ODE using the Runge-Kutta method

Greetings,

I've been working on a 2nd order ODE: y''(t) = -e^(3t)*y'(t) - y(t) + (5-2e^(-3t))*e^(-2t) +1

With initial conditions y(0) = 2 ; y'(0) = -2 where 0<t<1

I need to use a third order Runge Kutta method, which I can code for a 1st order ODE.

However given a 2nd order differential equation, I'm having difficulties implementing the ODE into my Runge Kutta code.

I've tried writing the 2nd order ODE in a linear system of 1st order ODEs but im still stuck.

By the way, i do not want to use ode45 or ode23 commands. Thank you for your feedback!

Here is what I've been working on so far:

function RK3 = func(a,b,n)
a = 0;
b = 1;
n = input('Enter the number of intervals:');
h = (b-a)/n;
y(1) = 2;
y(2) = -2;

for i=1:n
    t = (i-1)*h;
    k1 = f(t,y(:,i));
    k2 = f(t+0.5*h,y(:,i)+0.5*k1*h);
    k3 = f(t+h,y(:,i)-k1*h+(2*k2*h));
    y(i+1)= y(:,i)+h/6*(k1+4*k2+k3);  %approximated value of the ODE using RK3
end

for k=1:n+1
    t=a+(k-1)*h;
    exact(k)=exp(-2*t)+1;  %exact value of the ODE
end

error = max(abs(y-exact));  
max(exact)
max(y)
error 
end

function fty = f(t,y)  %matrix form of the 2nd order ODE
y = [0 1; -1 -exp(-3*t)];  
u = [0; 5-2*exp(-3*t)*exp(-2*t)+1];
fty = y + u;
end

Answer 1

[Alan Stevens]

More like this:

a = 0;
b = 1;
%n = input('Enter the number of intervals:');
n = 100;
h = (b-a)/n;
y = [2; -2];  
t = 0:h:n*h;  
for i=1:n
    k1 = f(t(i),y(:,i));
    k2 = f(t(i)+0.5*h,y(:,i)+0.5*k1*h);
    k3 = f(t(i)+h,y(:,i)-k1*h+2*k2*h);
    y(:,i+1)= y(:,i)+h/6*(k1+4*k2+k3);  %approximated value of the ODE using RK3
end

exact = exp(-2*t)+1;
disp(max(abs(exact-y(1,:))))
plot(t,y(1,:),'r',t,exact,'b--'),grid
xlabel('t'),ylabel('y')
legend('RK3','Exact')

function fty = f(t,y)  %matrix form of the 2nd order ODE
Y = [0 1; -1 -exp(-3*t)]*y;  
u = [0; (5-2*exp(-3*t))*exp(-2*t)+1]; 
fty = Y + u;
end

[Emir Alp Karslioglu]

This code works like a charm! Thank you so much for your assistance. I need to work more on matlix manipulation in matlab it seems. Much appreciated!