,differential equation for the 2nd order system

dynamical systemsordinary differential equations

I would like to know which differential equation I have to solve in order to get the answer to a unitary step like the one report in figure:
I want that this curve will be obtained to a solution of my system of ode (2nd order ODE)

tell me if I'm wrong:
starting from :

$m\frac{d^2y_1}{dt}+b\frac{dy_1}{dt}+k y_1= 0$

I can compute the substitution $\frac{dy}{dt} = y_2 $

so the system of first order equation became :
\begin{cases}
y'_1 = y_2 \\
y'_2 = -\frac{b}{m}y_2 -\frac{k}{m} \\
\end{cases}
right ?
how can I set the constant ? which initial value should I use ?

EDIT
Maybe I'm not well explain! I don't wanna know the output but which value and boundary condition I have to use in the system that I wrote in order to get the set of response in function of the parameter $\zeta$ (and also know what is $\zeta$) I know the soultion but I wanna solve the differential equations and get this set of curve !! Consider my question like: which is the function to integrate …. with wich value or like :"I have to give an exercise: using a euler method find the solution of a unit step response for different value of $\zeta$ and $\omega_o$ "

Thanks for your precious support … what is $x(t)$ ?

Best Answer

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2\zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation $$y''+2\zeta y'+y=x.$$ As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $t\ge 0$, and of course $x(t)=0$ for $t<0$. Then solve $$y''+2\zeta y'+y=1$$ for $t\ge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
    sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
    plt.plot(t,sol[:,0], label="$\zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

enter image description here

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