Question

Do the laplace transform of L\left\{t-e^{3t}\right\}

Laplace transform

Do the laplace transform of $$L\left\{t-e^{3t}\right\}$$

2021-09-20

Step 1
$$L\left\{t-e^{3t}\right\}$$
We know that the Laplace transformation of
$$\displaystyle{L}{\left({t}\right)}={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}\cdot{t}{\left.{d}{t}\right.}={\frac{{{1}}}{{{s}^{{2}}}}}$$
$$\displaystyle{L}{\left({e}^{{{a}{t}}}\right)}={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}\cdot{e}^{{{a}{t}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{s}-{a}}}},{s}{>}{a}$$
So $$\displaystyle\rightarrow{L}{\left({t}-{e}^{{{3}{t}}}\right)}$$
$$\displaystyle\Rightarrow{L}{\left({t}\right)}-{L}{\left({e}^{{{3}{t}}}\right)}$$ by the defination of laplace transformation
$$\displaystyle\Rightarrow{\frac{{{1}}}{{{s}^{{2}}}}}-{\frac{{{1}}}{{{s}-{3}}}}$$
$$\displaystyle\Rightarrow{\frac{{{\left({s}-{3}\right)}-{s}^{{2}}}}{{{s}^{{2}}{\left({s}-{3}\right)}}}}$$
$$\displaystyle\Rightarrow{L}{\left({t}-{e}^{{{3}{t}}}\right)}={\frac{{{\left({s}-{3}\right)}-{s}^{{2}}}}{{{s}^{{2}}{\left({s}-{3}\right)}}}}$$