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Expert-verified Found in: Page 416 ### 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 # In Problems 3-10, determine the Laplace transform of the given function.$\frac{\mathbf{4}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{13}\mathbf{s}\mathbf{+}\mathbf{19}}{\mathbf{\left(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{13}\mathbf{\right)}}$

Therefore, the solution is${\mathcal{L}}^{-1}\left\{\frac{4{s}^{2}+13s+19}{\left(s-1\right)\left({s}^{2}+4s+13\right)}\right\}\left(t\right)=2{e}^{t}+2{e}^{-2t}\mathrm{cos}3t+{e}^{-2t}\mathrm{sin}3t$

See the step by step solution

## Step 1: Given Information

The given value is$\frac{4{s}^{2}+13s+19}{\left(s-1\right)\left({s}^{2}+4s+13\right)}$

## Step 2: Use partial fractions

Make partial fraction of the given function, as:

$\begin{array}{c}\frac{4{s}^{2}+13s+19}{\left(s-1\right)\left({s}^{2}+4s+13\right)}=\frac{4{s}^{2}+13s+19}{\left(s-1\right)\left({\left(s+2\right)}^{2}+{3}^{2}\right)}\\ =\frac{A}{s-1}+\frac{B\left(s+2\right)+3C}{{\left(s+2\right)}^{2}+{3}^{2}}\end{array}$

On comparing the numerator of above equation we get:

$4{s}^{2}+13s+19=\left(A+B\right){s}^{2}+\left(4A+B+3C\right)s+13A-2B-3C$

On comparing the coefficients, we get the system

$\left\{\begin{array}{l}A+B=4\\ 4A+B+3C=13\\ 13A-2B-3C=19\end{array}$

On simplifying further, we get

$\left\{\begin{array}{l}A=4-B\\ C=B-1\\ -18B=-36\end{array}$

On solving the equations, we get

$\left\{\begin{array}{l}A=2\\ B=2\\ C=1\end{array}$

Thus, the partial fractions are obtained as:

$\frac{4{s}^{2}+13s+19}{\left(s-1\right)\left({s}^{2}+4s+13\right)}=\frac{2}{s-1}+\frac{2\left(s+2\right)+3}{{\left(s+2\right)}^{2}+{3}^{2}}$

## Step 3: Take Inverse Laplace transform

Take inverse Laplace transform using ${\mathcal{L}}^{-1}\left\{\frac{s-a}{{\left(s-a\right)}^{2}+{b}^{2}}\right\}\left(t\right)={e}^{at}\mathrm{cos}bt$and ${\mathcal{L}}^{-1}\left\{\frac{b}{{\left(s-a\right)}^{2}+{b}^{2}}\right\}\left(t\right)={e}^{at}\mathrm{sin}bt$as:

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

Hence, the required inverse Laplace transform is

${\mathcal{L}}^{-1}\left\{\frac{4{s}^{2}+13s+19}{\left(s-1\right)\left({s}^{2}+4s+13\right)}\right\}\left(t\right)=2{e}^{t}+2{e}^{-2t}\mathrm{cos}3t+{e}^{-2t}\mathrm{sin}3t$ ### Want to see more solutions like these? 