Graphing a 2nd order ODE

I am looking to graph the 2nd order ODE my''+cy'+ky = r(t). with m, c, and k, being adjustible variables. I have successfully made graphs where the m value where 1, yet I cannot figure out how to alter my code to make this work.

[VBBV]

Can you share your code ?

[Kyle Brahm]

Main Code:

function dydt = order2(t,y)

dydt = zeros(size(y));

a = 2.4; %coefficient for y. term

b = 1.44; %coefficient for y term

r = -10*t*exp(-1.2*t); %forcing function

dydt(1) = y(2);

dydt(2) = r -a*y(2) - b*y(1);

Command window Code:

tspan = [0 2];

y0 = [1,2];

[t,y]=ode45(@order2,tspan,y0);

plot(t,y(:,1))

Answer 1

[Sam Chak]

Hi @Kyle Brahm

I did something similar moment ago in a separate question. Perhaps you can refer to that and apply your own parameters:

https://www.mathworks.com/matlabcentral/answers/1703665-how-to-solve-this-second-differential-equation-mx-cx-kx-kx-3-f0cos-wt-i-am-stuck

function Demo_ODE

close all;

clear;

clc;

Tspan = [0 10]; % simulation time

y0 = [1; 2]; % initial values

[t, y] = ode45(@odefcn, Tspan, y0);

plot(t, y, 'linewidth', 1.5)

grid on

xlabel('Time, t [sec]')

ylabel({'$y_{1}\; and\; y_{2}$'}, 'Interpreter', 'latex')

legend({'$y_{1}$', '$y_{2}$'}, 'Interpreter', 'latex', 'location', 'best')

title('Time responses of the system')

end

function dydt = odefcn(t, Y)

dydt = zeros(2,1);

m = 1;

c = 2.4;

k = 1.44;

r = -10*t*exp(-1.2*t);

F = [0; r/m]; % input force vector

A = [0 1; -k/m -c/m]; % state matrix

dydt = A*Y + F; % state-space model

end

Result:

Answer 2

[Torsten]

function dydt = order2(t,y)
dydt = zeros(size(y)); 
a = 2.4; %coefficient for y. term 
b = 1.44; %coefficient for y term 
m = 2; %coefficient for y.. term
r = -10*t*exp(-1.2*t); %forcing function 
dydt(1) = y(2); 
dydt(2) = (r -a*y(2) - b*y(1))/m; 

Command window Code:

tspan = [0 2];
y0 = [1,2];
[t,y]=ode45(@order2,tspan,y0);
plot(t,y(:,1))