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Found in: Page 391

### 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

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# In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{5}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{\mathbit{y}}{\mathbf{=}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{:}}\phantom{\rule{0ex}{0ex}}{\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{2}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbit{w}}{\mathbit{h}}{\mathbit{e}}{\mathbit{r}}{\mathbit{e}}{\mathbf{}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}\left\{\begin{array}{l}0,0⩽t<1,\\ t,1

The solution of the given initial value problem using the method of Laplace transforms is

$y\left(t\right)=2{e}^{-2t}-2{e}^{-3t}+\left[\frac{1}{36}+\frac{1}{6}\left(t-1\right)-\frac{1}{4}{e}^{-2\left(t-1\right)}+\frac{2}{9}{e}^{-3\left(t-1\right)}\right]u\left(t-1\right)-\left[\frac{19}{36}+\frac{1}{6}\left(t-5\right)-\frac{7}{4}{e}^{-2\left(t-5\right)}+\frac{11}{9}{e}^{-3\left(t-5\right)}\right]u\left(t-5\right)]$

See the step by step solution

## Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equations into algebraic equations.T

${\mathbit{F}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbf{\int }}}_{{\mathbf{0}}}^{\mathbf{+}\mathbf{\infty }}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{·}{\mathbf{e}}^{\mathbf{-}\mathbf{s}\mathbf{·}\mathbf{t}}{\mathbf{d}}{\mathbit{t}}$

Where, F(s)= Laplace transformation

S= Complex number

t= real number>=0

t'= first derivative of teh function f(t)

## Step 2: Apply Laplace transform:

Given initial value problem

$y\text{'}\text{'}+5y\text{'}+6y=g\left(t\right)$

where Also

$g\left(t\right)=t,1

Using rectangular and unit function we can write

$g\left(t\right)=t\left[u\left(t-1\right)-u\left(t-5\right)\right]+1u\left(t-5\right)\phantom{\rule{0ex}{0ex}}=tu\left(t-1\right)-\left(t-1\right)u\left(t-5\right)\phantom{\rule{0ex}{0ex}}$$y\left(0\right)=0andy\text{'}\left(0\right)=2$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+5\mathcal{L}y\text{'}\left(s\right)+6\mathcal{L}y\left(s\right)=\mathcal{L}\left[g\left(t\right)\right]$

${s}^{2}\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+5s\mathcal{L}y\left(s\right)-5y\left(0\right)+6y\left(s\right)=\mathcal{L}\left[tu\left(t-1\right)-\left(t-1\right)u\left(t-5\right)\right]\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}y\left(s\right)+0-2+5s\mathcal{L}y\left(s\right)-0+6\mathcal{L}y\left(s\right)={e}^{-8}\left[\frac{1}{{s}^{2}}+\frac{1}{s}\right]-{e}^{-5s}\left[\frac{1}{{s}^{2}}+\frac{4}{s}\right]\left({s}^{2}+5s+6\right)\mathcal{L}y\left(s\right)-2\phantom{\rule{0ex}{0ex}}=\left[\frac{{e}^{-8}\left(1+s\right)}{{s}^{2}}\right]-\left[\frac{{e}^{-5s}\left(1+4s\right)}{{s}^{2}}\right]\phantom{\rule{0ex}{0ex}}$

$\mathcal{L}y\left(s\right)=\frac{2}{\left({s}^{2}+5s+6\right)}+\left[\frac{{e}^{-A}\left(1+s\right)}{{s}^{2}\left({s}^{2}+5s+6\right)}\right]-\left[\frac{{e}^{-5s}\left(1+1s\right)}{{s}^{2}\left({s}^{2}+5s+6\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{2}{s+2}-\frac{2}{s+3}+\left[\frac{{e}^{s}\left(1+s\right)}{{s}^{2}\left({s}^{2}+5s+6\right)}\right]-\left[\frac{{e}^{5s}\left(1+4s\right)}{{s}^{2}\left({s}^{2}+5s+6\right)}\right]\cdots \left(1\right)$

Using partial fraction

$\frac{1+s}{{s}^{2}\left({s}^{2}+5s+6\right)}=\frac{\frac{1}{36}}{s}+\frac{\frac{1}{6}}{{s}^{2}}-\frac{\frac{1}{4}}{s+2}+\frac{\frac{2}{9}}{s+3}\phantom{\rule{0ex}{0ex}}\frac{1+4s}{{s}^{2}\left({s}^{2}+5s+6\right)}=\frac{\frac{19}{36}}{s}+\frac{\frac{1}{6}}{{s}^{2}}-\frac{\frac{7}{4}}{s+2}+\frac{\frac{11}{9}}{s+3}$

Equation first becomes as

$\mathcal{L}y\left(s\right)=\frac{2}{s+2}-\frac{2}{s+3}+{e}^{-s}\left[\frac{\frac{1}{36}}{s}+\frac{\frac{1}{6}}{{s}^{2}}-\frac{\frac{1}{4}}{s+2}+\frac{\frac{2}{9}}{s+3}\right]-{e}^{-5s}\left[\frac{\frac{19}{36}}{s}+\frac{\frac{1}{6}}{{s}^{2}}-\frac{\frac{7}{4}}{s+2}+\frac{\frac{11}{9}}{s+3}\right]$

## Step 3: Take inverse Laplace transform we get

$y\left(t\right)=2{e}^{-2t}-2{e}^{-3t}+\left[\frac{1}{36}u\left(t-1\right)+\frac{1}{6}\left(t-1\right)u\left(t-1\right)-\frac{1}{4}{e}^{-2\left(t-1\right)}u\left(t-1\right)+\frac{2}{9}{e}^{-3\left(t-1\right)}u\left(t-1\right)\right]\phantom{\rule{0ex}{0ex}}-\left[\frac{19}{36}u\left(t-5\right)+\frac{1}{6}\left(t-5\right)u\left(t-5\right)-\frac{7}{4}{e}^{-2\left(t-5\right)}u\left(t-5\right)+\frac{11}{9}{e}^{-3\left(t-5\right)}u\left(t-5\right)\right]\phantom{\rule{0ex}{0ex}}=2{e}^{-2t}-2{e}^{-3t}+\left[\frac{1}{36}+\frac{1}{6}\left(t-1\right)-\frac{1}{4}{e}^{-2\left(t-1\right)}+\frac{2}{9}{e}^{-3\left(t-1\right)}\right]u\left(t-1\right)\phantom{\rule{0ex}{0ex}}-\left[\frac{19}{36}+\frac{1}{6}\left(t-5\right)-\frac{7}{4}{e}^{-2\left(t-5\right)}+\frac{11}{9}{e}^{-3\left(t-5\right)}\right]u\left(t-5\right)\phantom{\rule{0ex}{0ex}}$

Hence

$y\left(t\right)=2{e}^{-2t}-2{e}^{-3t}+\left[\frac{1}{36}+\frac{1}{6}\left(t-1\right)-\frac{1}{4}{e}^{-2\left(t-1\right)}+\frac{2}{9}{e}^{-3\left(t-1\right)}\right]u\left(t-1\right)-\left[\frac{19}{36}+\frac{1}{6}\left(t-5\right)-\frac{7}{4}{e}^{-2\left(t-5\right)}+\frac{11}{9}{e}^{-3\left(t-5\right)}\right]u\left(t-5\right)]$

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