# Dynamical system output - differential equations

I am applying at a Faculty for Electrical Engineering, and I have an entrance exam in two days. I have few exams for exercise, from previous years and I keep getting stuck in one particular type of problem:

"A dynamicaI system is described by a differential equation of the following form: $$y''(t) + 5y'(t)+4y(t) = u(t)$$ where t is the independent variable of time, $u(t)$ is the input signal into the system, and $y(t)$ is the system response."

I need to prove that if the system is excited with stepped amplitude signal of $1$, the output of the system in the steady state will reach a constant value of $0.25$.

Can anyone please suggest how can I solve this problem. I think I should use Laplace Transformation to solve it, but I do not know to how transform the above differential equation in some type that can be solved with Laplace. Also, i do not get where does the amplitude signal of $1$ is used.

Any help or hints are welcomed and appreciated, I need all the help I can get. Thank you.

• Maybe this would suit ELECTRICAL ENGINEERING.SE better? Aug 29 '17 at 9:21
• Thank you for the suggestion, I did not know there is a page for Electrical Engineering. I will try there. Aug 29 '17 at 9:33
• The question definitely belongs to math.SE. To solve this, why not use that every solution of the differential equation is $$y(t)=ae^{-t}+be^{-4t}+\frac13e^{-t}\int_0^tu(s)e^sds-\frac13e^{-4t}\int_0^tu(s)e^{4s}ds$$ for some constants $(a,b)$ depending on the initial conditions $(y(0),y'(0))$, and plug in the function $u(t)$ you are interested in, whatever it is? For instance, if $u(t)=1$ for every $t>0$, one gets a fully explicit formula for $y(t)$, from which the limit when $t\to\infty$ is direct...
– Did
Aug 29 '17 at 9:36
• An electrical engineer would probably do Laplace tranformation $$(s^2+5s+4)Y=\frac{1}{s}$$ then do the partial fractions, solve the system and ensure that the limit is as expected. Or maybe use the final value theorem instead if allowed.
– A.Γ.
Aug 29 '17 at 10:05
• I started solving it like that and ended with $$Y=\frac{1}{4}-\frac{1}{3}e^{-t}+\frac{1}{12}e^{-4t}$$ , and I am stuck again. Aug 29 '17 at 10:17

To answer your question you do not actually need to solve the differential equation. Indeed, in steady state conditions both $y´´$ and $y´$ will vanish: hence you are (asymptotically) left with the equation $$4 y(t) = 1$$ (as $u(t)$ is where the step input function comes into play) from which one concludes $$y = 1/4 = 0.25$$
• @KirylPesotski If you feed in the harmonic signal $e^{i\omega t}$ you will get your output $y_p(t)$ as a harmonic oscillation with no existing limit as $t\to\infty$. Your $y_s$ is not a steady state, but an amplitude (frequency response at $\omega$). I do not see your point.