Laplace Transformations can someone kindly help me with these few questions? :)
Find $L{e^tf(t)}$ in terms of $f*(s)$ and state a range of $s$ which this is defined.
I couldn't figure this out. You use the definition but i then get $e^t(1-s)f(t)$ and then i don't know what to do....
Using the Convolution Theorem, find the function f(t) satisfying the equation
$$f(t) = \int_0^t e^u f(t-u)\mathrm du + e^t$$
I know how to take the LT of both sides. I'm not sure how to figure out the LT of the integral though. Ive for f(s) = something + 1/(s-1) at the moment. 
 A: The solution to the first question follows easily from the definition: Can you bring $\int_0^\infty  {e^{ - st} e^t f(t)\,dt}$ into the form $\int_0^\infty  {e^{ - s't} f(t) \,dt}$? (Consider the hint you were already given above.)
The solution to the second question follows easily from the Convolution Theorem. You want to find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation
$$
f(t) = \int_0^t {e^u f(t - u)\,du}  + e^t, \;\; t \geq 0.
$$
Your approach is right.
Taking Laplace transform on both sides gives
$$
\hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1,
$$
where $\hat \varphi$ is the Laplace transform of the convolution
$$
(f*e^t )(t) = \int_0^t {e^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)e^{t - u}\,du}\bigg).
$$
(Note that the upper limit of the integral is $t$ in order to satisfy $t-u \geq 0$.) The Convolution Theorem states that the Laplace transform of a convolution is the product of the Laplace transforms of the individual functions. Hence,
$$
\hat \varphi(s) = \hat f(s) \frac{1}{{s - 1}}.
$$ 
Now you can solve for $\hat f(s)$, and, in turn by inversion (assuming that $s > 2$), find $f(t)$.
Note: After you find $f(t)$, verify that it satisfies the original equation. This is very easily done; indeed $f(t)$ is only slightly different from $e^t$ (in form).
EDIT (completing the solution; see Remark below): 
Solving for $\hat f(s)$ gives
$$
\hat f(s) = \frac{1}{{s - 2}}.
$$
Assuming that $s > 2$, it follows by inversion that $f(t)=e^{2t}$. Indeed, this $f$ satisfies the original equation; that is,
$$
e^{2t}  = \int_0^t {e^u e^{2(t - u)} \,du}  + e^t ,\;\; t \geq 0.
$$   
Remark. In view of the OP's comments below, it appears that the factor $u$ was forgotten in the convolution term of the original equation. A solution to the modified problem has now been posted.
A: Here is a detailed solution to the modified -- substantially more challenging -- problem (see the OP's comments below the previous answer; in particular, it is stated there that this is not homework). 
To find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation
$$
f(t) = \int_0^t {u e^u f(t - u)\,du}  + e^t, \;\; t \geq 0,
$$
begin, as in the previous answer, by writing
$$
\hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1,
$$
where this time $\hat \varphi$ is the Laplace transform of the convolution
$$
(f*te^t )(t) = \int_0^t {ue^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)(t-u)e^{t - u}\,du}\bigg).
$$
By the convolution theorem,
$$
\hat \varphi(s) = \hat f(s) \frac{1}{{(s - 1)^2 }}.
$$
It is worth noting that the term $1/(s-1)^2$ can be derived as follows, recalling that an exponential random variable with density function $\lambda e^{-\lambda t}$, $t \geq 0$, has mean equal to $1/\lambda$ (here $\lambda = s -1 > 0$):
$$
\int_0^\infty  {e^{ - st} te^t \,dt}  = \frac{1}{{s - 1}}\int_0^\infty  {t(s - 1)e^{ - (s - 1)t} \,dt}  = \frac{1}{{(s - 1)^2 }}.
$$
Solving for $\hat f(s)$ (using the above expression for $\hat \varphi(s)$) gives
$$
\hat f(s) = \frac{{s - 1}}{{s^2  - 2s}} = \frac{{(s - 2) + 1}}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{s}\frac{1}{{s - 2}}.
$$
Assuming that $s > 2$, it follows by inversion (and the Convolution Theorem) that 
$$
f(t) = 1 + (1 * e^{2t})(t),\;\; t \geq 0.
$$
(Indeed, note that $\int_0^\infty  {e^{ - st} 1\,dt}  = \frac{1}{s}$ and $\int_0^\infty  {e^{ - st} e^{2t} \,dt}  = \frac{1}{{s - 2}}$.)
Finally, from 
$$
(1 * e^{2t})(t) = \int_0^t {e^{2u} 1\,du} \,\bigg( = \int_0^t {1e^{2(t - u)} \,du} \bigg) ,
$$
it follows that
$$
f(t) = 1 + \frac{{e^{2t}  - 1}}{2} = \frac{{e^{2t}  + 1}}{2},\;\; t \geq 0.
$$
Indeed, this $f$ satisfies the original equation; that is, as one can easily verify, it holds
$$
\frac{{e^{2t}  + 1}}{2} = \int_0^t {ue^u \frac{{e^{2(t - u)}  + 1}}{2}\,du}  + e^t .
$$
EDIT (in response to the OP's comment below). While inverting $1/s$ gives $1$ and inverting $1/(s-2)$ gives $e^{2t}$, inverting $1/(s(s-2))$ does not give the product of $e^{2t}$ and $1$; rather, by the Convolution Theorem, it gives the convolution of $e^{2t}$ and $1$. Since $(1 * e^{2t})(t) = \frac{{e^{2t}  - 1}}{2}\,( = \int_0^t {e^{2u} \,du} )$,
$$
f(t) = 1 + \frac{{e^{2t}  - 1}}{2} = \frac{{e{}^{2t} + 1}}{2}
$$
(which was verified by substitution into the original equation).
However, as the OP observed, the solution can be obtained more elementarily by splitting $\hat f(s)$ into partial fractions. Specifically, 
$$
\hat f(s) = \frac{{s - 1}}{{s^2  - 2s}} = \frac{1}{{2s}} + \frac{1}{{2(s - 2)}},
$$
from which it follows, by inversion, that
$$
f(t) = \frac{1}{2} + \frac{1}{2}e^{2t}  = \frac{{e^{2t}  + 1}}{2}.
$$
