Understanding the First Order Prototype Solution in Time Domain

[Dustinsfl]

Consider the first order prototype,
\[
\frac{dy(t)}{dt} + \frac{1}{\tau}y(t) = f(t),\]
driven by a step function,
\[
f(t) = \mathcal{U}(t) =
\begin{cases}
1, & \text{if } t \geq 0\\
0, & \text{otherwise}
\end{cases}
\]
Find the solution in the time domain. That is, find the transient and steady state solution.

What does this mean? The second part is use a Laplace transform. I can find the solution that way no problem but not sure about what I need to do for finding the solution in the time domain.

 

[gtoguy97]

The solution in the time domain can be found by solving the differential equation. Since we have a first order differential equation, the solution can be found using the integrating factor method. We start by rewriting the differential equation as:$$\frac{dy}{dt} + \frac{1}{\tau}y=f(t)$$Now, we multiply both sides of the equation by the integrating factor $e^{\frac{t}{\tau}}$ to obtain:$$e^{\frac{t}{\tau}}\frac{dy}{dt} + e^{\frac{t}{\tau}}\frac{1}{\tau}y=e^{\frac{t}{\tau}}f(t)$$We can now integrate both sides of the equation with respect to $t$ to get:$$e^{\frac{t}{\tau}}y(t) = \int_{0}^{t}e^{\frac{x}{\tau}}f(x)dx + C_1$$where $C_1$ is an integration constant. The transient solution is given by the integral on the right hand side and the steady state solution is obtained when we let $t\rightarrow \infty$. In this case, the steady state solution is simply the value of the step function $f(t)$, that is, $f(t)=1$ for all $t\geq 0$.Therefore, the solution of the differential equation is given by:$$y(t) = e^{-\frac{t}{\tau}}\left(\int_{0}^{t}e^{\frac{x}{\tau}}dx + C_1\right)+ 1$$where $C_1$ is an integration constant.