Differential EquationInverse Transform Problems 2

Find $$\mathrm{F}(\mathrm{t})$$ if its Laplace transform is given by $$f(s)=\frac{s-1}{s^{2}+2 s+10}+\frac{1}{\sqrt{s-1}}$$

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

$$\mathcal {L}^{-1}\left\{\frac{s-1}{s^{2}+2 s+10}+\frac{1}{\sqrt{s-1}}\right\} $$

$$\mathcal {L}^{-1}\left\{\frac{s-1}{s^{2}+2 s+10}\right\}+\mathcal {L}^{-1}\left\{\frac{1}{\sqrt{s-1}}\right\} $$

$$\mathcal {L}^{-1}\left\{\frac{s-1}{s^{2}+2 s+1-1+10}\right\}+\mathcal {L}^{-1}\left\{\frac{1}{\sqrt{s-1}}\right\} $$

$$\mathcal{L}^{-1}\left\{\frac{s-1}{(s+1)^{2}+9}\right\}+\mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s-1}}\right\} $$

$$S \longrightarrow S-1 \quad \quad  S \longrightarrow S+1$$

$$e^{-t} \mathcal{L}^{-1}\left\{\frac{s-2}{s^{2}+9}\right\}+e^{t} \mathcal{L}^{-1}\left\{\frac{1}{\sqrt s}\right\} $$

$$\left.e^{-t}[\cos (3 t)-\frac{2}{3} \sin (3 t)\right]+e^{t} \frac{1}{\sqrt{\pi}} \cdot \frac{1}{\sqrt{t}} $$

$$=F(t)$$

Find $$F(t)$$ if $$L\{F(t)\}=f(s),$$ and $$\frac{d f}{d s}=\frac{-1}{s^{2}+4 s+13} $$

$$\frac{d f}{d s}=\mathcal {L}\{-t F(t)\} $$

$$\mathcal{L}\{-t F(t)\}=\frac{-1}{s^{2}+4 s+13} $$

$$\mathcal {L}^{-1} \mathcal{L}\{ F(t)\}=\mathcal {L}^{-1} \frac{1}{s^{2}+4 s+13} $$

$$t \cdot F(t)=\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+4 s+13}\right\} $$

$$=\mathcal {L}^{-t}\left\{\frac{1}{s^{2}+4 s+4-4+13}\right\} $$

$$=\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+4 s+4+9}\right\}=\mathcal {L}^{-1}\left\{\frac{1}{(s+2)^{2}+9}\right\} $$

$$S \rightarrow S-2$$

$$=e^{-2 t} \mathcal{L}^{-1}\left\{\frac{1}{s^{2}+9}\right\}=e^{-2 t} \cdot \frac{1}{3} \cdot \sin (3 t)$$

$$F(t)=\frac{e^{-2 t} \sin (3 t)}{3 t} $$

Evaluate $$L^{-1}\left\{\ln \left(1-\frac{2}{s}\right)\right\}=F(t)$$

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

$$f(s)=\ln \left(1-\frac{2}{s}\right)=\ln \left(\frac{s-2}{s}\right)=\ln (s-2)-\ln (s)$$

$$\ln \left(\frac{a}{b}\right)=\ln a-\ln b$$

$$f^{\prime}(s)=\frac{1}{s-2}-\frac{1}{s} $$

$$L[-t f(t)]=\frac{1}{s-2}-\frac{1}{s} $$

$$\mathcal {L}^{-1} L[-t F(t)]=\mathcal {L}^{-1}\left\{\frac{1}{s-2}-\frac{1}{s}\right\} $$

$$-t F(t)=\mathcal {L}^{-1}\left\{\frac{1}{s-2}-\frac{1}{s}\right\}=e^{2 t}-1$$

$$F(t)=\frac{e^{2 t}-1}{-t} $$