Solving a linear differential equation using Laplace transform

Question

I'm trying to solve the following problem

$x'+x=f(t)$, $x'(0)=1$

$f(x) = \begin{cases} 0, & \mbox{if } t\in [0,1[ \\ 1, & \mbox{if } t\in[1,+\infty[\end{cases}$

I have no doubts about how to take the Laplace transform of the 1st member of my equation. However I have no idea on how to treat the 2nd member $f(t)$? Can someone give me a hint? Then I can solve the equation and invert my Laplace transformation without doubts.

Answer 1

Your integral is $$\int_0^1 0dt+\int_1^\infty e^{-ts}dt = \dfrac{e^{-s}}{s}\,, s>0.$$

Or if you have learned basic manipulation of Laplace transform, then $$\mathcal{L}\{f(t)\} = \mathcal{L}\{u_1(t)\cdot 1\} = e^{-s}\mathcal{L}\{1\} = \dfrac{e^{-s}}{s}.$$

Answer 2

It looks like $f(t)=u(t-1)$ where $u$ is the unit step function, the laplace transform of $u(t-a)$ is $\frac {e^{-as}}{s}$