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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.${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{5}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{\mathbit{y}}{\mathbf{=}}{\mathbit{t}}{\mathbit{u}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{-}}{\mathbf{2}}{\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}{\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$

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

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

See the step by step solution

## Step 1: Define Laplace Transform

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

$y\text{'}\text{'}+5y\text{'}+6y=tu\left(t-2\right)$

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

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+5\mathcal{L}y\text{'}\left(s\right)+6\mathcal{L}y\left(s\right)=\mathcal{L}\left[tu\left(t-2\right)\right]$

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

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

Using partial fraction

$\frac{1+2s}{{s}^{2}\left({s}^{2}+5s+6\right)}=\frac{\frac{7}{36}}{s}+\frac{\frac{1}{6}}{{s}^{2}}-\frac{\frac{3}{4}}{s+2}+\frac{\frac{5}{9}}{\left(s+3\right)}$

Equation first becomes as

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

Taking inverse Laplace transform we get

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

Hence,

$y\left(t\right)={e}^{-2t}-{e}^{-3t}+\left[\frac{7}{36}+\frac{t-2}{6}-\frac{3}{4}{e}^{-2\left(t-2\right)}+\frac{5}{9}{e}^{-3\left(t-2\right)}\right]u\left(t-2\right)$ ### Want to see more solutions like these? 