F(s) = L{ƒ(1)} = ¸ƒ(t)x™“dt (t)e-ª

(a) $b\frac{d^{3}f\left(t\right)}{d{t}^3}-\left(1-b\right)\frac{df\left(t\right)}{dt}$

We know that Laplace transformation of $F^n\left(t\right)$ is

$$\begin{gathered} L\left(f^n\left(t\right)\right)=s^{n}L\left\{f\left(t\right)\right\}-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-....-f^{n-1}\left(0\right) \\ =s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-....-f^{n-1}\left(0\right) \left[\because L\left\{f\left(t\right)\right\}=F\left(s\right)\right] \end{gathered}$$

From the linear property of Laplace transformation we have 

$$\begin{gathered} L\left(cf\left(t\right)\right)=cL\left(f\left(t\right)\right) \\ =cF\left(s\right) \end{gathered}$$

$$\begin{gathered} \therefore b\frac{d^{3}f\left(t\right)}{d{t}^3}-\left(1-b\right)\frac{df\left(t\right)}{dt} \\ =L\left(b\frac{d^{3}f\left(t\right)}{d{t}^3}-\left(1-b\right)\frac{df\left(t\right)}{dt}\right) \\ =bL\left(\frac{d^{3}f\left(t\right)}{d{t}^3}\right)-\left(1-b\right)L\left(\frac{df\left(t\right)}{dt}\right) \\ =b\left\{s^{3}L\left(f\left(t\right)\right)-s^{2}f\text{'}\text{'}\left(0\right)-sf\text{'}\left(0\right)-f\left(0\right)\right\}-\left(1-b\right)\left\{sL\left(f\left(t\right)\right)-f\left(0\right)\right\} \\ =b{s}^{3}L\left(f\left(t\right)\right)-b{s}^{2}f\text{'}\text{'}\left(0\right)-bsf\text{'}\left(0\right)-bf\left(0\right)-sL\left(f\left(t\right)\right)+f\left(0\right)+bsL\left(f\left(t\right)\right)-bf\left(0\right) \\ =b{s}^{3}F\left(s\right)-b{s}^{2}f\text{'}\text{'}\left(0\right)-bsf\text{'}\left(0\right)-bf\left(0\right)-sF\left(s\right)+f\left(0\right)+bsF\left(s\right)-bf\left(0\right) \\ =\left(b{s}^3-s+bs\right)F\left(s\right)-b{s}^{2}f\text{'}\text{'}\left(0\right)-bsf\text{'}\left(0\right)-f\left(0\right)\left(2b+1\right) \end{gathered}$$

Hence $L\left(b\frac{d^{3}f\left(t\right)}{d{t}^3}-\left(1-b\right)\frac{df\left(t\right)}{dt}\right)=\left(b{s}^3-s+bs\right)F\left(s\right)-b{s}^{2}f\text{'}\text{'}\left(0\right)-bsf\text{'}\left(0\right)-f\left(0\right)\left(2b+1\right)$

(b) $a{e}^{at}\sin \left(at\right)-b{e}^{-bt}\cos \left(bt\right)$

From the linear property of Laplace transformation we have 

$$\begin{gathered} L\left(cf\left(t\right)\right)=cL\left(f\left(t\right)\right) \\ =cF\left(s\right) \end{gathered}$$

From first shifting property we have 

$L\left(e^{at}f\left(t\right)\right)=F\left(s-a\right)$

$$\begin{gathered} \therefore a{e}^{at}\sin \left(at\right)-b{e}^{-bt}\cos \left(bt\right) \\ =L\left(a{e}^{at}\sin \left(at\right)-b{e}^{-bt}\cos \left(bt\right)\right) \\ =aL\left(e^{at}\sin \left(at\right)\right)-bL\left(e^{-bt}\cos \left(bt\right)\right) \\ =aF\left(s-a\right)-bF\left(s+b\right) \\ L\left(\sin \left(at\right)\right)=F\left(s\right) \\ =\frac{a}{s^2+a^2} \\ \therefore F\left(s-a\right)=\frac{a}{{\left(s-a\right)}^2+a^2} \\ L\left(\cos \left(bt\right)\right)=F\left(s\right) \\ =\frac{s}{s^2+b^2} \\ \therefore F\left(s-b\right)=\frac{\left(s-b\right)}{{\left(s-b\right)}^2+b^2} \\ \therefore a{e}^{at}\sin \left(at\right)-b{e}^{-bt}\cos \left(bt\right) \\ =\frac{a^2}{{\left(s-a\right)}^2+a^2}-\frac{b\left(s-b\right)}{{\left(s-b\right)}^2+b^2} \end{gathered}$$

Hence $L\left(a{e}^{at}\sin \left(at\right)-b{e}^{-bt}\cos \left(bt\right)\right)=\frac{a^2}{{\left(s-a\right)}^2+a^2}-\frac{b\left(s-b\right)}{{\left(s-b\right)}^2+b^2}$

(c) $f\left(kt-m\right)$

From second shifting property we have 

$$\begin{gathered} L\left(f\left(t\right)\right)=F\left(s\right) \\ G\left(t\right)=F\left(t-a\right) t>0 \\ =0 t>0 \\ L\left(G\left(t\right)\right)=e^{-as}F\left(s\right) \end{gathered}$$

From change of scale property

$L\left(f\left(at\right)\right)=\frac{1}{a}F\left(\frac{s}{a}\right)$

Using the first shifting property then change of scale property we have 

$$\begin{gathered} L\left(f\left(kt-m\right)\right)=e^{-ms}L\left(f\left(kt\right)\right) \\ =\frac{e^{-ms}}{k}F\left(\frac{s}{k}\right) \end{gathered}$$

Hence $L\left(f\left(kt-m\right)\right) =\frac{e^{-ms}}{k}F\left(\frac{s}{k}\right)$