`Iff(y)=e^y ,g(y)=y;y>0,a n dF(t)=int_0^t f(t-y)g(y) dy,t h e n` a) `F(t)=e^t-(1+t)` b) `F(t)=t e^t` c) `F(t)=t e^(-t)` (d) `F(t)=1-e^t(1+t)`

Options

A. `F(t)=e^(t)-(1+t)`
B. `F(t)=te^(t)`
C. `F(t)=te^(-t)`
D. `F(t)=1-e^(t)(1+t)`

Answer

A

Explanation

We have `f(y)=e^(y),g(y)=y,ygt0`
`F(t)=int_(0)^(t)f(t-y)g(y)dy`
`=int_(0)^(t)e^(t-y) y dy`
`=e^(t)int_(0)^(t) e^(-y) y dy`
`=e^(t) ([-ye^(-y)]_(0)^(t)+int_(0)^(t) e^(-y) dy)`
`=e^(t) (-te^(-t) -[e^(-y)]_(0)^(t))`
`=e^(t) (-te^(-t)-e^(-1) +1)`
`=e^(t)-(1+t)`