• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q30E

Expert-verified
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

# In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{10}}{\mathbit{y}}{\mathbf{=}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{;}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{,}}{\mathbf{}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbit{w}}{\mathbit{h}}{\mathbit{e}}{\mathbit{r}}{\mathbit{e}}{\mathbf{}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}\left\{\begin{array}{l}10,0⩽t⩽10,\\ 20,10

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

$y\left(t\right)=-2{e}^{-t}\mathrm{cos}3t-\frac{2}{3}{e}^{-t}\mathrm{sin}3t+1-{e}^{-\left(t-10\right)}\mathrm{cos}3\left(t-10\right)-\frac{1}{3}{e}^{-\left(t-10\right)}\mathrm{sin}3\left(t-10\right)+u\left(t-10\right)\phantom{\rule{0ex}{0ex}}+2{e}^{-\left(t-20\right)}\mathrm{cos}3\left(t-20\right)+\frac{2}{3}{e}^{-\left(t-20\right)}\mathrm{sin}3\left(t-20\right)-2u\left(t-20\right)$
See the step by step solution

## Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equation into differential equations into algebraic equations. the formula for laplace transform is

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

Where, F(s)= Laplace transform

S= Complex number

t= real number>=0

t’ = first derivative of the function f(t)

## Step 2: Apply Laplace transform

Given initial value problem

$y\text{'}\text{'}+2y\text{'}+10y=g\left(t\right)$

where.$y\left(0\right)=-1andy\text{'}\left(0\right)=0$ Also

$g\left(t\right)=10,0⩽t⩽10\phantom{\rule{0ex}{0ex}}=20,10⩽t⩽20$

Using rectangular and unit function we can write

$g\left(t\right)=10-10u\left(t-10\right)+20u\left(t-10\right)-20u\left(t-20\right)\phantom{\rule{0ex}{0ex}}=10+10u\left(t-10\right)-20u\left(t-20\right)$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+2\mathcal{L}y\text{'}\left(s\right)+10\mathcal{L}y\left(s\right)=\mathcal{L}\left[g\left(l\right)\right]$

${s}^{2}\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+10y\left(s\right)=\mathcal{L}\mid 10+10u\left(t-10\right)-20u\left(t-20\right)\right]\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}y\left(s\right)+s-0+2s\mathcal{L}y\left(s\right)+2+10\mathcal{L}y\left(s\right)=\frac{10}{s}+\frac{10c}{s}-\frac{20{e}^{10s}}{s}\phantom{\rule{0ex}{0ex}}\left({s}^{2}+2s+10\right)\mathcal{L}y\left(s\right)+\left(s+2\right)=\frac{10}{s}+\frac{10{t}^{-11\right)}}{s}-\frac{20{e}^{-10s}}{s}$

$\mathcal{L}y\left(s\right)=-\frac{s+2}{\left({s}^{2}+2s+10\right)}+\frac{10}{s\left({s}^{2}+2s+10\right)}+\frac{10{e}^{-10s}}{s\left({s}^{2}+2s+10\right)}-\frac{20{e}^{-10s}}{s\left({s}^{2}+2s+10\right)}\phantom{\rule{0ex}{0ex}}=-\frac{s+1}{{\left(s+1\right)}^{2}+9}-\frac{1}{{\left(s+1\right)}^{2}+9}+\frac{10}{s\left({s}^{2}+2s+10\right)}+\frac{10{e}^{-10s}}{s\left({s}^{2}+2s+10\right)}-\frac{20{e}^{-10s}}{s\left({s}^{2}+2s+10\right)}\phantom{\rule{0ex}{0ex}}$

Using partial fraction

$\frac{10}{s\left({s}^{2}+2s+10\right)}=\frac{-s-2}{{s}^{2}+2s+10}+\frac{1}{s}$

Equation first becomes as

$\mathcal{L}y\left(s\right)=-\frac{s+1}{{\left(s+1\right)}^{2}+9}-\frac{1}{{\left(s+1\right)}^{2}+9}+\frac{-s-2}{{s}^{2}+2s+10}+\frac{1}{s}+\left[\frac{-s-2}{{s}^{2}+2s+10}+\frac{1}{s}\right]{e}^{-10s}-\left[\frac{-s-2}{{s}^{2}+2s+10}+\frac{1}{s}\right]2{e}^{-20s}$

## Step 3: Take inverse Laplace transform we get

$y\left(t\right)=-{e}^{-t}\mathrm{cos}3t-\frac{1}{3}{e}^{-t}\mathrm{sin}3t-{e}^{-t}\mathrm{cos}3t-\frac{1}{3}{e}^{-t}\mathrm{sin}3t+1-{e}^{-\left(t-10\right)}\mathrm{cos}3\left(t-10\right)-\frac{1}{3}{e}^{-\left(t-10\right)}\mathrm{sin}3\left(t-10\right)+u\left(t-10\right)\phantom{\rule{0ex}{0ex}}+2{e}^{-\left(t-20\right)}\mathrm{cos}3\left(t-20\right)+\frac{2}{3}{e}^{-\left(t-20\right)}\mathrm{sin}3\left(t-20\right)-2u\left(t-20\right)\phantom{\rule{0ex}{0ex}}=-2{e}^{-t}\mathrm{cos}3t-\frac{2}{3}{e}^{-t}\mathrm{sin}3t+1-{e}^{-\left(t-10\right)}\mathrm{cos}3\left(t-10\right)-\frac{1}{3}{e}^{-\left(t-10\right)}\mathrm{sin}3\left(t-10\right)+u\left(t-10\right)\phantom{\rule{0ex}{0ex}}+2{e}^{-\left(t-20\right)}\mathrm{cos}3\left(t-20\right)+\frac{2}{3}{e}^{-\left(t-20\right)}\mathrm{sin}3\left(t-20\right)-2u\left(t-20\right)\phantom{\rule{0ex}{0ex}}$

hence

$y\left(t\right)=-2{e}^{-t}\mathrm{cos}3t-\frac{2}{3}{e}^{-t}\mathrm{sin}3t+1-{e}^{-\left(t-10\right)}\mathrm{cos}3\left(t-10\right)-\frac{1}{3}{e}^{-\left(t-10\right)}\mathrm{sin}3\left(t-10\right)+u\left(t-10\right)\phantom{\rule{0ex}{0ex}}+2{e}^{-\left(t-20\right)}\mathrm{cos}3\left(t-20\right)+\frac{2}{3}{e}^{-\left(t-20\right)}\mathrm{sin}3\left(t-20\right)-2u\left(t-20\right)$