I have the following problem :
$$m\ddot{x} + c\dot{x} + kx = f_f\delta(t-t_0) + f_c \sin(\omega t) + f_h \theta (2t_0-t)$$
where $x(t)$ is a function of time, $t>0$ and $t_0>0$ and where $\delta(t)$ is the Dirac distribution and $\theta(t)$ is the Heaviside step function. The initial conditions are $$x(0) = \dot{x}(0) = 0$$
Because of the Dirac distribution and the Heaviside function in the right hand side and because of the initial conditions, I tried by Laplace Transform... But this is what I get :
\begin{align} m\left[s^2 X(s) -x(0) - \dot{x}(0)\right] + c\left[s X(s) -x(0)\right] + k X(s) = f_f e^{-t_0 s} + f_c\cdot\frac{\omega}{s^2+\omega^2} + f_h \cdot \theta(2t_0)\frac{1-e^{-2t_0 s}}{s} \\ (ms^2+cs+k)X(s) = f_f e^{-t_0 s} + f_c\cdot\frac{\omega}{s^2+\omega^2} + f_h \cdot \theta(2t_0)\frac{1-e^{-2t_0 s}}{s} \\ X(s) = f_f \frac{e^{-t_0 s}}{ms^2+cs+k} + f_c\cdot\frac{\omega}{s^2+\omega^2}\cdot\frac{1}{ms^2+cs+k} + f_h \cdot \theta(2t_0)\frac{1-e^{-2t_0 s}}{s(ms^2+cs+k)} \end{align}
The "only" thing remaining is to do the inverse transformation. I think I could manage the first term and most probably the last if I succeed in the first... But for the second term multiplied by $f_c$? With convolution? Is there no simpler solution to this ?
(I used Wolfram alpha to find the laplace transform of $\theta (2t_0-t)$, I hope the result is right)
First of all, your LTs are all technically correct, although you do not need the $\theta(t_0)$ term, as that is $1$ because $t_0\gt 0$.
One way to get the ILT of the second term is to use partial fraction decomposition. That is, express
$$\frac{\omega}{s^2+\omega^2} \frac1{m s^2+c s+k} = \frac{A s+B}{s^2+\omega^2}+\frac{C s+D}{m s^2+c s+k}$$
and solve for $A$, $B$, $C$, and $D$. It is a slog, but the result is
$$\left ( \begin{array} \\A\\B\\C\\D\end{array}\right )=\left(\begin{array}-\frac{c \omega}{c^2 \omega^2+k^2-2 k m \omega^2+m^2 \omega^4}\\\frac{\omega \left(k-m \omega^2\right)}{c^2 \omega^2+k^2-2 k m \omega^2+m^2 \omega^4}\\\frac{c m \omega}{c^2 \omega^2+k^2-2 k m \omega^2+m^2 \omega^4}\\\frac{\omega \left(c^2-k m+m^2 \omega^2\right)}{c^2 \omega^2+k^2-2 k m \omega^2+m^2 \omega^4}\end{array}\right)$$
You should be able to find the ILT now given what you said.
Alternatively, you can use the residue theorem if you are familiar with it.