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Expert-verified Found in: Page 350 ### 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 ${1}{-}{14}$ , solve the given initial value problem using the method of Laplace transforms.${\mathbf{3}}{\mathbf{.}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{9}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}{}{}{}{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{,}}{}{}{}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{6}}$

The Initial value for $y\text{'}\text{'}+6y\text{'}+9y=0$ is $y\left(t\right)=-{e}^{-3t}+3t{e}^{-3t}$

See the step by step solution

## Step 1: Define the Laplace Transform

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$

## Step 2: Determine the initial value of Laplace transform

Applying the Laplace transform and using its linearity we get

$\mathcal{L}\left\{y\text{'}\text{'}+y\text{'}+9y\right\}=0\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}+6\mathcal{L}\left\{y\text{'}\right\}+9\mathcal{L}\left\{y\right\}=0\phantom{\rule{0ex}{0ex}}\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+6\left[sY\left(s\right)-y\left(0\right)\right]+9Y\left(s\right)=0$

Solve for the Laplace transform as:

${s}^{2}Y\left(s\right)+s-6+6\left[sY\left(s\right)+1\right]+9Y\left(s\right)=0\phantom{\rule{0ex}{0ex}}{s}^{2}Y\left(s\right)+6sY\left(s\right)+9Y\left(s\right)=-s\phantom{\rule{0ex}{0ex}}\left({s}^{2}+6s+9\right)Y\left(s\right)=-s\phantom{\rule{0ex}{0ex}}Y\left(s\right)=\frac{-s}{{s}^{2}+6s+9}$

Using partial fractions solve as:

$\frac{-s}{{s}^{2}+6s+9}=\frac{-s}{{\left(s+3\right)}^{2}}=\frac{A}{s+3}+\frac{B}{{\left(s+3\right)}^{2}}\phantom{\rule{0ex}{0ex}}-s=A\left(s+3\right)+B$

Using $s=-3,0$ , , respectively, gives

$s=-3:3=B⇒B=3\phantom{\rule{0ex}{0ex}}s=0:0=3A+B⇒A=-1$

Therefore

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

Using the inverse Laplace transform we obtain the solution of given differential equation

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

Therefore,

$y\left(t\right)=-{e}^{-3t}+3t{e}^{-3t}$

Therefore, the initial value for $y\text{'}\text{'}+6y\text{'}+9y=0$ is $y\left(t\right)=-{e}^{-3t}+3t{e}^{-3t}$ ### Want to see more solutions like these? 