Consider the initial value problem:
$$x''(t) +3x'(t)+2x(t)= t + (ae^{-(t-a)}-t) H(t-a)$$ $$x(0)=0$$ $$x'(0)=0$$ where $a>0$ is constant and $H$ is the Heaviside function.
Let $f(t)=t + (ae^{-(t-a)}-t) H(t-a)$. Applying the Laplace transformation with $ \mathcal{L \{x(t) \}}=X(s)$ and $\mathcal{L \{f(t) \} } =F(s)$, and using the initial conditions, and after some calculation we get:
$X(s) =\dfrac{F(s)}{(s+1)(s+2)}$, from which we get $x(t) = \mathcal{L}^{-1} \left\{\dfrac{F(s)}{(s+1)(s+2)} \right\}$.
And so it is the case that: $$x(t) = \mathcal{L}^{-1} \left\{\frac{F(s)}{s+1} \right\} -\mathcal{L}^{-1} \left\{\frac{F(s)}{s+2} \right\} = e^{-t}*f(t) - e^{-2t}*f(t)= e^{-t}(1-e^{-t})*f(t)$$
where $*$ is the operator of convolution.
How can I continue now to do as few calculations as possible?
Thanks in advance!
Edit: I am interested also if there is a simpler way (using Laplace Transformation). Tell me.
The best way is to compute the convolution integral in the $s$-plane. Not in the time plane. Use the identity: $$\mathcal{L\{f(t)\ast g(t)\}}=\mathcal{L \left\{\int_o^tf(t-\tau)g(\tau)d\tau \right\} } =F(s)G(s),$$ where $F(s)$ and $G(s)$ are the Laplace transforms of $f(t)$ and $g(t)$ respectively.