Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the convolution theorem to find the inverse Laplace transform of the given function.$\frac{\mathbf{s}}{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}}\left[Hint: \frac{{\displaystyle s}}{{\displaystyle s-1}}=1+\frac{{\displaystyle 1}}{{\displaystyle s-1}}\right]$

Step-by-Step Solution

Answer

The inverse Laplace transform for the given function by using the convolution theorem is.$y\left(t\right)=\frac{2e^{-2t}}{3}+\frac{e^t}{3}$

Step 1: Define convolution theorem

Let and be piecewise continuous on $[0,\infty )$and of exponential orderand set,$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{f}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\text{ and }}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{g}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$ then,

$\mathcal{L}\mathbf{\{}\mathit{f}\mathbf{\ast }\mathit{g}\mathbf{\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{,}$or

${\mathcal{L}}^{-1}\left\{F\right(s\left)G\right(s\left)\right\}\left(t\right)=(f\ast g)\left(t\right)$

Step 2: Use the convolution theorem to obtain the inverse Laplace transform

Consider the given equation,

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

Let,

$\begin{array}{c}y\left(s\right)=\frac{s}{(s-1)(s+2)}\\ =\frac{1}{s+2}[1+\frac{1}{s-1}]\\ =\frac{1}{s+2}+\frac{1}{(s-1)(s+2)}\end{array}$

Take inverse Laplace transform on both sides,

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

Hence, the convolution formula is,${\mathcal{L}}^{-1}\left[f\right(s)\cdot g(s\left)\right]=f\star g={\int }_0^{t}f(t-v)g\left(v\right)dv$, where

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

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

Thus, the equationcan be written as,

$\begin{array}{c}y\left(t\right)=e^{-2t}+{\int }_0^t{e}^{t-v}\cdot e^{-2v}dv\\ =e^{-2t}+e^t{\int }_0^t{e}^{-3v}dv\\ =e^{-2t}+e^t{\left[\frac{e^{-3v}}{-3}\right]}_0^t\\ =e^{-2t}-\frac{e^t}{3}[e^{-3t}-1]\end{array}$

$\begin{array}{c}y\left(t\right)=e^{-2t}-\frac{e^{-2t}}{3}+\frac{e^t}{3}\\ =\frac{2e^{-2t}}{3}+\frac{e^t}{3}\end{array}$

Therefore, the inverse Laplace transform for the given function is. $y\left(t\right)=\frac{2e^{-2t}}{3}+\frac{e^t}{3}$