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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Determine the inverse Laplace transform of the given function.$\frac{\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{6}}$

The solution is

${\mathcal{L}}^{-1}\left\{\frac{2\left(s-2\right)+3}{{\left(s-2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}\right\}\left(t\right)=2{e}^{2t}\mathrm{cos}\sqrt{2}t+\frac{3}{\sqrt{2}}{e}^{2t}\mathrm{sin}\sqrt{2}t$

See the step by step solution

## Step 1: Given Information

The given function value in s domain is $\frac{2s-1}{{s}^{2}-4s+6}$

## Step 2: Use partial fractions

Factorize the denominator of the given function as

$\begin{array}{c}{s}^{2}-4s+6={s}^{2}-4s+4+2\\ ={\left(s-1\right)}^{2}+{\left(\sqrt{2}\right)}^{2}\end{array}$

The function becomes.$\frac{2s-1}{{\left(s-2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}$Decompose the function as:

$\begin{array}{c}\frac{2s-1}{{\left(s-2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}=\frac{2\left(s-2\right)+3}{{\left(s-2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}\\ =\frac{2\left(s-2\right)}{{\left(s-2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}+\frac{3}{\sqrt{2}}\frac{\sqrt{2}}{{\left(s-2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}\end{array}$

Take inverse Laplace transform using and as:

Therefore,the required inverse Laplace transform is: