Runge-Kutta solving 2nd ODE

Hey guys, given that y''+3y'+2y=2e^(-3t)， with y(0)=0,y'(0)=1.

let x1=y, x2=y'

dx1=y'=x2------->(1)

dx2=y''=2e^(3t)-3x2-2x1------->(2)

but there's somethings confusing me, what should i write for F1 since it doesn't have idea what it y'.

I'm new to MATLAB, so i really have no idea what i'm doing.

my TA ask me what is y' but i have no idea what i'm gonna tell him what's y'.

function RungeKuttaMethod
a = 0;                  %time interval we are solving over
b = 4;
N = 1000;               %number of steps
%t = zeros(1, N);       
%y = zeros(1, N);
%y(0) = 0;             %initial value of y.
%y'(0) = 1;             %initial value of y'.
h = (b - a)/N;          %step size

 
F1 = @(t,y)y' ;
F2 =  @(t, y)2*exp(-3*t)-3*y'-2*y;   %thing we are solving.
for i = 1:(N-1)
 
K11 = h*(F1(t(i),y(i)));   
 
K21 = h*(F2(t(i), y(i))); 
 
K12 = h*(F1(t(i)+0.5*h,y(i)+0.5*K21));
 
K22 = h*(F2(t(i) + 0.5*h, y(i) + 0.5*K11));
 
K13 = h*(F1(t(i)+0.5*h,y(i)+0.5*K22));
 
K23 = h*(F2(t(i) + 0.5*h, y(i) + 0.5*K12));
 
K14 = h*(F1(t(i)+h,y(i)+K22));
 
K24 = h*(F2(t(i) + h, y(i) + K13));
 
y1(i+1) = y(i) + (K11 + 2*K12 + 2*K13 + K14)/6;
 
y2(i+1) = y(i) + (K21 + 2*K22 + 2*K23 + K24)/6;      
 
t(i+1) = a + i*h;
end
 
plot(t, y1)
hold on
plot(t,y2)
 
 
end
 
 
 

Answer 1

[darova]

See attached code. Try to write your code as clear and simple as it possible, because it is very easy to make a mistake

Answer 2

[Jan]

ODE integrators work with functions of 1st order only. If you have an equation of 2nd order, you can convert it easily to a system of equations of the 1st order. Instead of calculating y'' you create a system of 2 equations. Then y' is simply the 2nd component of the input, exactly as you have written already:

let x1=y, x2=y'

THis is actuallly trivial. If you have a problem to see it, stop working and drink a cup of coffee.