how can i solve second order ODE with RK-4 without using a built in function in matlab?

this is the equation :

(d^2 y)/(dt^2 ) + 3 (dy)/dt - 30t - 10 = 0

this is my code :

h=0.01;          %step size (changable according to the proplem)
t=0:h:1;        %the tdomain range
y = [1;-3]; %intial condition for first equation in form of matrix (changable according to the proplem)
for i = 1:length(t)-1
    
    K11 = RK4(t(i), y1(:, i));
    K12 = RK4(t(i) + h/2, y(:, i) + h*K11/2);
    K13 = RK4(t(i) + h/2, y(:, i) + h*K12/2);
    K14 = RK4(t(i) + h,   y(:, i) + h*K13);
    y(:, i+1) = y(:, i) + h/6*(K11 + 2*K12 + 2*K13 + K14);
    
end

function dy1 = RK4(t, y)
    f11 = y(2);
    f12 = -3*y(1)+30*t+10;
    dy1 = [f11;f12];
    %change the matrix due to your intital conditins 
                      %and the equation in your proplem
end

is this right?

[Mohammad Adeeb]

.

[Mohammad Adeeb]

.

Answer 1

[Alan Stevens]

This

K11 = RK4(t(i), y1(:, i));

should be |

K11 = RK4(t(i), y(:, i));

and this

f12 = -3*y(1)+30*t+10;

should be |

f12 = -3*y(2)+30*t+10;

[Mohammad Adeeb]

the first sugeestion is o; , but the second one , we should i use 2 instead of 1?

[Alan Stevens]

Think of your equation as

% dv/dt + 3*v - 30*t - 10 = 0  where  v = dy/dt
%
% The y you pass to your RK4 function is [y; v]
% i.e. y(1) = y and y(2) = v
% and you should return [v; dv/dt]
% so v = y(2) and dv/dt = -3*y(2)+30*t+10;
 

[Mohammad Adeeb]

but in this way i didnt use the conditon y(1)

[Alan Stevens]

Look at your original equation. It doesn't contain y on its own.

[Mohammad Adeeb]

ok , thank u