How to solve $t-2f(t) = \int_0^t(e^\tau- e^{-\tau})f(t-\tau)d\tau$ I want to solve this equation. It reminds me something about Laplace transform.
I am sure that I must use it order to solve it.
$$t-2f(t) = \int_0^t(e^\tau- e^{-\tau})f(t-\tau)d\tau$$
How to do it?
 A: The right hand side of your equation is a convolution of two functions.  The Laplace transform of the convolution, $f_1\star f_2$, of $f_1$ and $f_2$ is:
$$\tag{1}
{\cal L}\bigl[ (f_1 \star f_2)(t)\bigr]= {\cal L}\Bigl[\,\int_0^t f_1(\tau) f_2(t-\tau) \,d\tau\Bigr]=F_1(s)F_2(s),$$
where $F_1(s)$ is the Laplace transform of $f_1(t)$ and $F_2(s)$ is the Laplace transform of $f_2(t)$.
Taking the Laplace transform of both sides of your equation gives
$$
\cal L \bigl[ t-2f(t)\bigr] = \cal L  \bigl[\, (e^t-e^{-t})\star f(t)\,\bigr].
$$
By  equation (1):
$$
\cal L [ \,t -2 f(t)\,\bigr] =  \, \cal L[e^t - e^{-t}]\,  \cdot
\cal L\bigl[f(t)\bigr].
$$
By linearity of the Laplace transform:
$$\cal L [ \,t\,] -2\cal L\bigl[ f(t)\bigr] = \bigl(\, \cal L[e^t]-\cal L[e^{-t}]\,\bigr) \cdot
\cal L\bigl[f(t)\bigr].$$
So, denoting the Laplace transform of $f(t)$ by $F(s)$:
$$
\tag{2}{1\over s^2}-2 F(s) =\Bigl[ { 1\over s-1}-{1\over s+1}\Bigr]F(s).
$$
Solving equation (2) for $F(s)$ gives
$$
F(s)={s^2-1\over 2s^4}={1\over 2s^2}-{1\over 2s^4}.
$$ Applying the inverse Laplace transform now gives
$$
f(t)={1\over2} t-{1\over12}t^3.
$$
