If f(y)=e^y, g(y)=y; y0 and F(t) = _0^t f(t-y)g(y),dy, then:

Question

If $f(y)=e^y$, $g(y)=y$; $y>0$ and $F(t) = \displaystyle\int_0^t f(t-y)g(y)\,dy$, then:

Accepted Answer

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$$

$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$

$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$

$$ =  - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$$

$$ =  - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$$

$$ =  - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$$

$$ = {e^t} - \left( {1 + t} \right)$$

Solutions · 1