Solving a constant coefficients ODE where the input $x(t)$ is not continuous

Question

I’m an electrical engineering student and just started taking a course in signals and systems. We were given the following constant coefficients ODE.

$$6y’’(t) -45y’(t) +21y(t) = 2x(t) $$

1. Find the Characteristic Polynomial
2. find the Homogenous solution
3. find the conditions for the zir solution given $y(0)=5$ and $y’(0)=31/2$
4. find the conditions for the zsr solution given $x(t) = u(t)$ (u(t) is the heaviside step function.)
   
   what do you think they mean by conditions? Do you think they wanted us to solve the system?
   I know how to find the homogenous and ZIR solution of the system, starting with guessing $y=e^{st}$, calculating characteristic polynomial and finding the constants by substituting in the initial conditions. $$y_{zir} = 2e^{7t} +3e^{\frac12 t}$$ when it came to finding the ZSR solution I was stumped - had the input been continuous I’d have guessed a similar form for the solution. I found a video where an equation approximating this one had been solved using the laplace transform though we would get to it later in the course.
   Is there a different method?
   If there isn’t how would you solve this equation with laplace as well as use ZSR/ZIR?

$$\begin{cases} 6y’’_{ZSR}(t) -45y’_{ZSR}(t) +21 = x(t) \\ y’_{ZSR}(0)=0 \quad y_{ZSR}(0) = 0 \end{cases}$$

Answer 1

For the ZSR, you need some particular solution combined with a homogeneous solution to cancel out that particular solution's value at $t = 0$. Since the input is discontinuous, let's consider the regions on each side of the continuity separately. For $t > 0$, it's pretty clear $f(t) = 2/21$ is a particular solution, while for $t < 0$, $f(t) = 0$ is. Then we need to find $A$ and $B$ such that $A e^{7t} + B e^{t/2} + f(t)$ has zero value and derivative at $0$. For $t > 0$, this gives $A + B +2/21= 0$ and $7A + B/2 = 0$, which solves to $A = -2/315$ and $B = 4/45$. For $t < 0$, this gives $A = B = 0$. Writing this in terms of the unit step function gives $$y_{ZSR} = u(t)\left[\frac{2}{21} + \frac{2}{273}e^{7t} - \frac{4}{39}e^{t/2}\right].$$ You get the same answer doing it through Laplace transforms. Zero state means the boundary terms vanish, so you can use partial fractions to get $$ 6s^2 Y-45s Y + 21s = \frac{2}{s}\Longrightarrow Y = \frac{1}{3s(s-7)(2s-1)} =\frac{2}{21s} + \frac{2}{273(s-7)}-\frac{8}{39(2s+1)}, $$ which obviously inverts to the $y_{ZSR}$ given above.