Laplace transform of this problem's solution

Question

Hi have to determinate the Laplace Transform of u where u is the solution to this problem:

I really have no idea on how to manage this mess.

Answer 1

For any real numbers $a\in\mathbb{R}$, let $H(t-a)$ be the shifted unit step function, i.e. $$ H(t-a) = \begin{cases} \, 1 & \ \ \textrm{ if }t\ge a, \\ \, 0 & \ \ \textrm{ if }t<a. \end{cases} $$ One can rewrite the given equation as follows \begin{align*} u(t) - \int_t^\infty u(s)\, ds & = \chi_{[1,2]}(t) \\ u(t) - \left(\int_0^\infty u(s)\, ds - \int_0^t u(s)\, ds\right) & = H(t-1) - H(t-2) \\ u(t) - 1 + \int_0^t u(s)\, ds & = H(t-1) - H(t-2). \end{align*} Denote the Laplace transform of $u$, $\mathcal{L}(u)(s)$ by $U(s)$. Taking Laplace transform of the final equation gives \begin{align*} U(s) - \frac{1}{s} + \frac{U(s)}{s} & = \frac{e^{-s}}{s} - \frac{e^{-2s}}{s} \\ U(s)\left(1+ \frac{1}{s}\right) & = \frac{1}{s}[e^{-s} - e^{-2s} + 1] \\ \implies U(s) & = \frac{e^{-s} - e^{-2s} + 1}{s+1}. \end{align*}

Answer 2

Well you have an equation in terms of $u$. Slap on an $e^{-st}$ and integrate. We have $$\int_0^\infty u(t)e^{-st}dt - \int_0^\infty e^{-st}\int_t^\infty u(t')dt'dt = \int_0^\infty e^{-st}\chi_{[1,2]}(t)dt.$$ The RHS is $$\int_0^\infty e^{-st}\chi_{[1,2]}(t)dt = \int_1^2e^{-st}dt = \frac{1}{s}(e^{-s}-e^{-2s}) = C(s)/s.$$

The first term on the LHS is the Laplace transform $\mathcal L_u(s).$ The second term can be integrated by parts $$ \int_0^\infty e^{-st}F(t)dt = -\left.F(t)\frac{1}{s}e^{-st}\right|_0^\infty + \int_0^\infty\frac{1}{s}e^{-st}F'(t)dt$$ where $F(t) = \int_t^\infty u(t')dt'.$ We have $$F(0) = \int_0^\infty u(t')dt' = 1$$ by definition, $ \lim_{t\to \infty} F(t)=0$ by the convergence of the integral,and $F'(t) = -u(t)$ by the FTC so we get $$ \int_0^\infty e^{-st}F(t)dt = \frac{1}{s} - \int_0^\infty \frac{1}{s}e^{-st}u(t)dt = \frac{1}{s}-\frac{1}{s}\mathcal L_u(s).$$

Plugging this all into the initial equation, $$ (1+\frac{1}{s})\mathcal L_u(s) - \frac{1}{s} = C(s)/s$$ so we get $$ \mathcal L_u(s) = \frac{1}{1+s}(C(s)+1)$$ where $C(s) = e^{-s}-e^{-2s}.$