Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

 $\frac{\mathbf{14}}{\mathbf{(}\mathbf{s}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{5}\mathbf{)}}$

Step-by-Step Solution

Answer

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

 $y\left(t\right) = 2e^{5t} - 2e^{-2t}$

Step 1: Define convolution theorem

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

 then,

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

or

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

Step 2: Determine inverse Laplace transform for the given function

Consider the function,

$\frac{14}{(s + 2)(s - 5)}$

Let,

$y\left(s\right)=\frac{14}{(s + 2)(s - 5)}$

Take inverse Laplace transform on both sides,

$L^{-1}\left[y\right(s\left)\right] = L^{-1}\frac{14}{(s + 2)(s - 5)}$

Thus, use the convolution formula, $(f * g)\left(t\right)={\int }_0^{t}f(t - v)g\left(v\right)dv$, where

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

 $g\left(s\right) =\frac{1}{s-5}$and $f\left(t\right) =e^{5t}$

Hence, the equation becomes,

$$\begin{gathered} y\left(t\right) = \frac{1}{14} {\int }_0^t{e}^{-2\left(t-v\right)}{e}^{5v}dv \\ =14e^{-2t}{\int }_0^t{e}^{7v}dv \\ =14e^{-2t}{\left[\frac{e^{7v}}{7}\right]}_0^t \\ =2e^{-2t}\left[e^{7t}-1\right] \end{gathered}$$

$y\left(t\right)=2e^{5t}-2e^{-2t}$

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

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