Solve the following with Laplace transform $y''-y'=f(t)$ $$\begin{cases} 
     y''-y'=f(t) \\
      y(0)=1 \\
     y'(0)=1\\
   \end{cases}
$$
$f(t)$ is
$$f(t)= \begin{cases} 
      -1 & 0\leq t < \pi \\
      cos(t) & \pi\leq t < 2\pi \\
      1 & t \geq 2\pi \\
   \end{cases}
$$
what I need to do now?


*

*Convert $f(t)$ to one line function, if yes I would like to see.

*Force on LHS and RHS Laplace transform?

Thanks.
 A: You can write $f(t)$ as:
$$-u(t) + u(t- \pi)(\cos t +1) + u(t-2\pi)(1- \cos t)$$
The Laplace Transform of the LHS is:
$$\mathcal{L}\{y''- y'\} = (s^2 y(s) - sy(0) - y'(0)) - (sy(s) - y(0))$$
with: $y(0)=1, y'(0)=1$
Can you continue?
Update
We now have:
$(s^2y(s) -s -1) -(s y(s)-1) = \left(-\dfrac{1}{s}\right) + \left(\dfrac{e^{-\pi s}}{s} - \dfrac{e^{-\pi s} s}{1 + s^2}\right) +\left(\dfrac{e^{-2 \pi s}}{s} - \dfrac{e^{-2 \pi s} s}{1 + s^2}\right)$
This reduces to:
$$y(s) = \dfrac{1}{s(s-1)}\left(s + \left(-\dfrac{1}{s}\right) + \left(\dfrac{e^{-\pi s}}{s} - \dfrac{e^{-\pi s} s}{1 + s^2}\right) +\left(\dfrac{e^{-2 \pi s}}{s} - \dfrac{e^{-2 \pi s} s}{1 + s^2}\right) \right)$$
An example of an inverse Laplace transform is:
$$\mathscr{L}^{-1}\left(\dfrac{1}{s(s-1)} \dfrac{e^{-\pi s}}{s} \right) = (-1 -t + \pi + e^{-\pi + t}) U(t -\pi)$$
A: You equate the laplace transforms of left and right hand sides, where the laplace transform of the right hand side equals $-\int_0^\pi e^{-st} dt + \int_\pi^{2\pi} \cos t e^{-st} dt + \int_{2\pi}^\infty e^{-st},$ and the Laplace transform of the LHS is $L(y)$ times a polynomial. Then solve for $L(y),$ and invert the Laplace transform.
A: The function $f(t)$ is continuous in $\pi$ and $2\pi$, so you can write, in Laplace:
$$F(s)=-\frac{1}{s}(1-e^{-\pi t})+\frac{s}{s^2+1}(e^{-\pi t}-e^{-2\pi t})+\frac{1}{s}(e^{-2\pi})$$
Obtaining this with the Laplace time translation property.
Now you have to put this input in the function $Y(s)$ obtained by the system (just a multiplication), and you will obtain the result for $Y(s)$. Obviously, to have the time version of this result, you have just to anti transform.
