Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

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 1: 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

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

Step 2: 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

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

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

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" style="max-width: none; vertical-align: -4px;" >