Question

Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{\mathbf{1}}{\mathbf{(}\mathbf{s}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{)}}$

Short answer

The inverse Laplace transform for the given function using the convolution theorem is.

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

Step-by-step solution

Define convolution theorem

Let $\mathit{f}\left(t\right)$and $\mathit{g}\left(t\right)$be piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{\infty }\mathbf{)}$and of exponential order $\mathit{\alpha }$and set,

$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{L}\mathbf{}\left\{f\right\}\left(s\right)\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{G}\left(s\right)\mathbf{=}\mathit{L}\left\{g\right\}\left(s\right)$then,

$\mathit{L}\left(f*g\right)\left(s\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{}\mathit{G}\left(s\right)$

or

${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left(f\left(s\right)g\left(s\right)\left(t\right)\right)\mathbf{=}\mathbf{}\left(f*g\right)\left(t\right)$

Find the inverse Laplace transform by using the convolution theorem

Consider the given function,$\frac{1}{\left(s+1\right)\left(s+2\right)}$

Let,$F\left(s\right)=\frac{1}{\left(s+1\right)}$

$G\left(s\right)=\frac{1}{\left(s+2\right)}$

Since, we know

role="math" localid="1665059297605" $f\left(t\right)=L^{-1}\left\{\frac{1}{s+1}\right\}\left(t\right)=e^{-t}$

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

Apply inverse Laplace transform and use convolution theorem to obtain,

$$\begin{gathered} L^{-1}\left\{\frac{1}{(s+1)(s+2)}\right\}\left(t\right)=L^{-1}\{F(s\left)G\right(s\left)\right\}\left(t\right) \\ =g\left(t\right)*g\left(t\right) \\ =e^{-t}{e}^{-2t} \end{gathered}$$

Use the convolution formula,

$(f*g)\left(t\right)={\int }_0^{t}f(t-v)g\left(v\right)dv$

$$\begin{gathered} L{}^{-1}\left\{\frac{1}{(s+1)(s+2)}\right\}\left(t\right)={\int }_0^t{e}^{-\left(t-v\right)}{e}^{-2v}dv \\ =e^{-t}{\int }_0^t{e}^{-v}dv \\ =e^{-t}{\left[-e^{-v}\right]}_0^t \\ =e^{-t}(1-e^{-t}) \end{gathered}$$

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

Therefore, the inverse Laplace transform for the given function is.

$L{}^{-1}\left\{\frac{1}{(s+1)(s+2)}\right\}\left(t\right)=e^{-t}(1-e^{-t})$" width="9" height="19" role="math">