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Prove that if $L^{-1}\{f(s)\}=F(t)$ then $L^{-1}\{f(k s)\}=\frac{1}{k} F\left(\frac{t}{k}\right), k>0$

$\mathcal{L} L^{-1}\{F(k s)\}=\mathcal {L} \frac{1}{k} F\left(\frac{t}{k}\right)$

$f(k s)=\frac{1}{k} \mathcal {L} F(t / k)$

$f(s)=\int_{0}^{\infty} e^{-s t} F(t) d t=L[F(t)]$

$f\left({ks}\right)=\int_{0}^{\infty} e^{-k s t} F(t) d t$

put $u=k t \rightarrow d u=k d t$

$=\int_{0}^{\infty} e^{-s u} F\left(\frac{u}{k}\right) \frac{d u}{k}=\frac{1}{k} L\left[F\left(\frac{u}{k}\right)\right]$

replace u by t

$f(k s)=\frac{1}{k} L\left[F\left(\frac{t}{k}\right)\right] \quad, k>0$

Show that if $F(t)$ is of class $A,$ than $f^{(n)}(s)=L\left\{(-t)^{n} F(t)\right\}$
Where $f(s)=L\{F(t)\}$

$f(s)=\mathcal {L}\{F(t)\}=\int_{0}^{\infty} e^{-s t} F(t) d t$

$f^{\prime}(s)=\int_{0}^{\infty} -t \cdot e^{-s t} F(t) d t=\mathcal{L}\{-t F(t)\}$

$f^{\prime \prime}(S)=\int_{0}^{\infty} t^{2}\cdot e^{-s t} F(t) d t=\mathcal{L}\left\{t^{2} F(t)\right\}$

$f^{\prime \prime}(s)=\int_{0}^{\infty}-t^{3} e^{-s t} F(t) d t=L\left\{-t^{3} F(t)\right\}$

$f^{n}(s)=\int_{0}^{\infty}(-t)^{n} e^{-s t} F(t) d t=\mathcal{L} \left\{(-t)^n F(t) dt\right\}$

If $F(s)=L[f(t)]$ exists for $s>a \geq 0$ and if $c$ is a constant
then $L\left[\mathrm{e}^{c t} f(t)\right]=f(s-c)$ and $\mathrm{e}^{c t} f(t)=L^{-1}[F(s-c)]$

$\mathcal {L}^{-1} L\left[e^{c t} f(t)\right]=\mathcal {L}^{-1} F(s-c)$

$e^{c t} f(t)=L^{-1}[F(s-c)]$

$L\left\{e^{c t} f(t)\right\}=\int_{0}^{\infty} {e^{-s t}} {e}^{c t} f(t) d t$

$=\int_{0}^{\infty} e^{-(s-c) t} f(t) d t$

$L\left\{e^{c t} f(t)\right\}=F(s-c) \quad \quad s>a+c$

$\mathcal {L}^{-1} L\left\{e^{c t} f(t)\right\}=\mathcal {L}^{-1}F(s-c)$

$e^{c t} f(t)=\mathcal{L}^{-1}[F(s-c)]$