Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{3}}\left(s^2+1\right)}$

Step-by-Step Solution

Answer

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

 $y\left(t\right)=\frac{1}{2}\left[t^2-2+2\cos t\right]$

Step 1: Define convolution theorem

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

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

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

Step 2: Find the inverse Laplace transform for the given function

Consider the function,

$\frac{1}{s^3\left(s^2+1\right)}$

Let, $y\left(s\right)=\frac{1}{s^3\left(s^2+1\right)}$

Take inverse Laplace transform on both sides,

${\mathcal{L}}^{-1}\left[y\right(s\left)\right]={\mathcal{L}}^{-1}\left[\frac{1}{s^3\left(s^2+1\right)}\right]$

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

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

$g\left(s\right)=\frac{1}{s^3} and g\left(t\right)=\frac{1}{2}{t}^2$

Hence, the equation 1 becomes,

$y\left(t\right)={\int }_0^t\sin (t-v)·\frac{{\displaystyle 1}}{{\displaystyle 2}}{v}^{2}dv$

Thus, by using parts rule,

$$\begin{gathered} y\left(t\right)=\frac{1}{2}{\left[v^2\cos (t-v)-{\int }_0^{t}2v\cos (t-v)dv\right]}_0^t \\ =\frac{1}{2}{\left[v^2\cos (t-v)-2[-v\sin (t-v)+\cos (t-v\left)\right]\right]}_0^t \\ =\frac{1}{2}\left[t^2\cos 0-2[-t\sin 0+\cos 0-0+2[0+\cos t\left]\right]\right] \\ =\frac{1}{2}\left[t^2-2+2\cos t\right] \end{gathered}$$

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

$y\left(t\right)=\frac{1}{2}\left[t^2-2+2\cos t\right]$