Question

Determine f (t) = {4sin (2t) − 6e4t}.

Answer 1

Laplace Transformation

Question. Find $\mathcal{L}\{\mathrm{F(t)\}}$, where

$\quad\quad \mathrm{F(t)=4\:sin2t-6\:e^{4t}}$.

Formulae.

$\quad\quad \mathcal{L}\mathrm{(e^{at})=\frac{1}{s-a},\:s>a}$

$\quad\quad \mathcal{L}\mathrm{(sinat)=\frac{a}{s^{2}+a^{2}}}$

Solution.

Now, $\mathcal{L}\mathrm{\{F(t)\}}$

$\quad=\mathcal{L}\mathrm{(4\:sin2t-6\:e^{4t})}$

$\quad=\mathcal{L}\mathrm{(4\:sin2t)}-\mathcal{L}\mathrm{(6\:e^{4t})}$

$\quad\mathrm{=4}\mathcal{L}\mathrm{(sin2t)-6}\mathcal{L}\mathrm{(e^{4t})}$

$\quad\mathrm{=4.\frac{2}{s^{2}+2^{2}}-6.\frac{1}{s-4}}$

$\quad\mathrm{=\frac{8}{s^{2}+4}-\frac{6}{s-4}}$

$\Rightarrow \boxed{\color{red}{\mathcal{L}\mathrm{(4\:sin2t-6\:e^{4t})=\frac{8}{s^{2}+4}-\frac{6}{s-4}}}}$

This is the required Laplace transform.

Definition.

Let $\color{blue}{\mathrm{F(t)}}$ be a transform of real variable $\color{blue}{\mathrm{t}}$ defined in $\color{blue}{\mathrm{(-\infty,\:\infty)}}$ such that $\color{blue}{\mathrm{F(t)=0}}$ for all $\color{blue}{\mathrm{t<0}}$. The Laplace transform of $\color{blue}{\mathrm{F(t)}}$ is denoted by $\color{blue}{\mathcal{L}\mathrm{\{F(t)\}=f(s)}}$ or $\color{blue}{\mathrm{f(p)}}$ defined by

$\quad \color{blue}{\mathrm{f(s)=}\mathcal{L}\mathrm{\{F(t)\}=\int_{0}^{\infty}e^{-st}F(t)dt}}$,

provided the improper or the infinite integral exists.

The parameter $\color{blue}{\mathrm{s}}$ or $\color{blue}{\mathrm{p}}$ may be real or complex.