Initial Value Problem of a Differential Equation Involving a Step Function

Question

We were given the problem

$$y'+ay=h_1 \quad y(0)=0$$ $$h_1(x) = \begin{cases} 1, & \text{if } 0 < x \leq 1 \\ 0, & \text{if } x> 1 \end{cases}$$

and tasked with finding the function $y$. We were told that one method we may use involves Green's Function, but we are allowed to use whatever method we choose.

What I have attempted

From having already calculated the Laplace transformation of $h_1$ previously to be $H(x)=\frac{1-e^{-s}}{s}$, I then set $$ \mathscr{L}[y](s)=Y(s)$$ $$ \mathscr{L}[y'](s)=sY(s)$$ giving me the transformed problem: $$sY(s)+aY(s)=\frac{1-e^{-s}}{s}$$ $$Y(s)=\frac{1-e^{-s}}{s(s+a)}$$ which looked really complicated to find the inverse of, so I decided there was probably an easier way.

I then tried to use Green's function (which I don't wholly understand): $$Y(s)=G(s)H(s), \quad G(s)=\frac{1}{\chi}=\frac{1}{s+a}=\mathscr{L}[e^{-ax}](s)$$ $$\Rightarrow g(x)=e^{-ax} $$ $$y(x)=(g*h_1)(x)=\int_0^xg(x-\tau)h_1(\tau)d\tau=\int_0^xe^{-a(x-\tau)}h_1d\tau$$

At this point I don't know if I'm making things any easier. Especially because I have again arrived at $Y(s)=\frac{1-e^{-s}}{s(s+a)}$. I don't know how to place $h_1$ into the equation. I know that I can also write it using the Heaviside function as $$1-u_1=1-u(t-1)$$ which gives me $$\int_0^xe^{-a(x-\tau)}(1-u(\tau-1))d\tau$$

Am I on the right track? Is there a better approach?

Answer 1

This isn't really an answer to your question, but the inverse transform of $$Y(s) = \frac{1 - e^{-s}}{s(s+a)}$$ isn't too bad. We see $$\frac{1}{s(s+a)} = \frac{1}a \left(\frac 1 s - \frac 1 {s+a} \right)$$ so $$Y(s) =\frac{1}{a}\left(\frac 1 s - \frac 1 {s+a}- \frac {e^{-s}} s + \frac {e^{-s}} {s+a} \right) $$ which gives $$y(t) = \frac{1}{a}\left( 1 -e^{-at} - u(t-1)(1 - e^{-a(t-1)} )\right)$$ where $u(t)$ is the unit step function: $$u(t) = \left\{ \begin{matrix} 0, & t < 0, \\ 1 & t \ge 0. \end{matrix} \right.$$ I'd stick with the Laplace transform method.