Numerical solution for a 2nd order ode

Hi everyone, I'm struggling to find the numerical solution for the following equation and conditions:

y'' + (0.2)*y' - y - y^3 = (0.3)*cos(wt) y(0) = 0 y'(0) = 0

I need to plot a numerical solution from ti = 0 to tf = 100 for w = 0.8:0.1:1.5. I am really new to matlab so fair warning, here's what I've got so far:

syms y t;
a = y;
b = Dy;
Db = (0.3)*cos(w*t)-(0.2)*b + a - a^3;

figure;
hold on
for x = 0.8:0.1:1.5
    ezplot(subs(Db, 'w', x), [0 100])
end
hold off
axis ([0 100 0 3])
title 'Numerical solutions of D2y + (0.2).*Dy - y - y.^3 = (0.3).*cos(w.*t) when w = 0.8, 0.9,..., 1.4, 1.5'
legend('w = 0.8', 'w = 0.9', 'w = 1.0', 'w = 1.1', 'w = 1.2', 'w = 1.3', 'w = 1.4', 'w = 1.5')

I'd appreciate any guidance or feedback, thank you!

[Jan]

"Numerical" solution means, that you do not use "syms".

Start with converting the 2nd order equation to a system of equations of the 1st order.

Answer 1

[Torsten]

Rewrite your ODE as

y1'=y2
y2'=-0.2*y2+y1+y1^3+0.3*cos(w*t)

and use ODE45 to solve.

Look at the examples provided under

https://de.mathworks.com/help/matlab/ref/ode45.html

Best wishes

Torsten.