2nd Order Differential equation with Unit Step and initial conditions

Question

I have some work but I can't get it to work quite yet, I need to solve this so I can plot it versus time.
The function is $90y''+30y'+1000(y-z(t))=0$,
$$z(t)= \begin{cases}0 & 0\leq t\leq5/9 \\ -\frac18 & 5/9<t\leq40/9\\0 & 40/9\leq t\leq 20/3\end{cases}$$ with $y(0)=0$ and $y'(0)=0$. I know $z(t)$ is a heaviside function but I am having difficulty moving forward. I've tried to simplify the main function to $90/1000y''+30/1000y'+y=z(t)$, and the heaviside notation I have is $z(t)=-\frac18u_{5/9}(t)$.
Help would be greatly appreciated. I apologize for the poor notation, I am new here and I do not know how to properly put in my functions.

Answer 1

If you just want to plot it, you can also get the values numerically

def z(t): return -1/8 if 5<9*t<=40 else 0

t = np.linspace(0,7,500)
y = odeint(lambda y,t: [ y[1], -(3*y[1]+100*(y[0]-z(t)))/9 ], [0,0], t)
plt.plot(t,y)

to get

Answer 2

If you're interested in an analytical solution . . .

Your ODE is \begin{align} \ddot{y}+\alpha \dot y+\beta y=\beta z, \end{align} where I've defined \begin{align} \alpha&=\frac{3}{9}\\ \beta&=\frac{100}{9} \end{align} for brevity. Substituting \begin{align} x(t)=\gamma y+\dot y \end{align} where $\gamma$ satisfies \begin{align} \gamma^2-\alpha \gamma+\beta=0 \end{align} reduces your ODE to \begin{align} \dot x+\frac{\beta}{\gamma}x=\beta z. \end{align} This has the solution \begin{align} x=e^{-\beta t/\gamma}\left(k_1+\int\beta e^{\beta t/\gamma}z\mathrm dt\right). \end{align} For the right integral \begin{align} -\frac{1}{8}\int \beta e^{\beta t/\gamma}(\mathrm H_{5/9}-\mathrm H_{40/9})\mathrm dt, \end{align} using integrating by parts \begin{align} -\frac{\gamma}{8}\left((e^{\beta t/\gamma}-e^{5\beta/9\gamma})\mathrm H_{5/9}-(e^{\beta t/\gamma}-e^{40 \beta/9\gamma})\mathrm H_{40/9}\right). \end{align} So \begin{align} \dot y+\gamma y=k_1e^{-\beta t/\gamma}+\frac{\gamma}{8}\left((\mathrm{exp}\left[\frac{\beta}{\gamma}\left(\frac{5}{9}-t\right)\right]-1)\mathrm H_{5/9}-(\mathrm{exp}\left[\frac{\beta}{\gamma}\left(\frac{40}{9}-t\right)\right]-1)\mathrm H_{40/9}\right)\equiv f(t) \end{align} and \begin{align} y=e^{-\gamma t}\left[k_2+\int e^{\gamma t}f(t)\mathrm dt\right], \end{align} where you could solve the right integral using integration by parts again.