Q7.4 - 3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{s}\mathbf{+}\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{10}}$.

Step-by-Step Solution

Answer

The inverse laplace transform of the given function is ${\mathrm{e}}^{-\mathrm{t}}\cos 3\mathrm{t}$.

Determining the inverse laplace transform

• For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
• Consider a function F(s) , if there is a function f(t) that is continuous on $[0,\infty )$ and satisfies $\mathcal{L}\left\{f\right\}=F$ then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation 
•  $f={\mathcal{L}}^{-1}\left\{F\right\}$
•  ${\mathcal{L}}^{-1}\left\{\frac{n!}{{(s-a)}^{n+1}}\right\}=e^{at}{t}^n,n=1,2,\dots$

Find inverse laplace transform for the given function

The given function is $\frac{s+1}{s^2+2s+10}$.

Find the inverse laplace transform of $\frac{s+1}{s^2+2s+10}$using ${\mathcal{L}}^{-1}\left\{\frac{s-a}{{(s-a)}^2+{\left(b\right)}^2}\right\}=e^{at}\cos bt$ as:

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{s+1}{{(s+1)}^2+{\left(3\right)}^2}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{s-(-1)}{\left[s-{(-1)}^2+{\left(3\right)}^2\right.}\right\}\\ & =& e^{-t}\cos 3t\end{array}$

Therefore, the inverse laplace transform of the given function is $e^{-t}\cos 3t$.