Respuesta :
Positive integer and a is a real constant. 12. f(t) = teat os [tex]$F(s) =\frac{1}{(s-a)^2} \Rightarrow L f(t)=F(s)=\frac{1}{(s-a)^2}$[/tex]
[tex]$f(t)=t e^{a t}$[/tex]
[tex]$\begin{aligned}& \quad \because L[t(t)]=F(s)=\int_0^{\infty} f(t) e^{-s t} d t \\& F(s)=\int_0^{\infty} t e^{a t} e^{-s t} d t \\& F(s)=\int_0^{\infty} t e^{-t(s-a)} \cdot d t \\& \because \int u v d t=u \int v-\int\left[\frac{d(u)}{d t} \times \int v d t\right] d t\end{aligned}$[/tex]
whir u and v is the function of t.
[tex]$ \therefore F(s)=\left\{\frac{t e^{-f(s-a)}}{-(s-a)}\right\}_0^{\infty}-\int_0^{\infty} \frac{1 \times e^{-t(s-a)}}{-(s-a)} d t $[/tex]
[tex]$F(s) =\frac{-1}{(s-a)}\{0-0\}+\frac{1}{s-a} \int_0^{\infty} e^{-t(s-a)} d t $[/tex]
[tex]$ =0+\frac{1}{(s-a)} \int_0^{\infty} e^{-t(s-a)} d t $[/tex]
[tex]$ =\frac{1}{(s-a)} \times\left\{\frac{e^{-t(s-a)}}{-(s-a)}\right\}_0^{\infty} $[/tex]
[tex]$ =\frac{-1}{(s-a)^2}\left\{e^{-\infty}-e^0\right\} $[/tex]
[tex]$ =\frac{-1}{(s-a)^2}\{0-1\} $[/tex]
[tex]$F(s) =\frac{1}{(s-a)^2} \Rightarrow L f(t)=F(s)=\frac{1}{(s-a)^2}$[/tex]
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