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### 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 ${\mathbf{1}}{\mathbf{-}}{\mathbf{14}}$, solve the given initial value problem using the method of Laplace transforms${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{7}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{10}}{\mathbf{y}}{\mathbf{=}}{\mathbf{9}}{\mathbf{cost}}{\mathbf{+}}{\mathbf{7}}{\mathbf{sint}}{\mathbf{;}}{}{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{5}}{\mathbf{,}}{}{}{}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{-}}{\mathbf{4}}$

The initial value for $y\text{'}\text{'}-7y\text{'}+10y=9\mathrm{cos}t+7\mathrm{sin}t$ is $y\left(t\right)=-4{e}^{5t}+8{e}^{2t}+\mathrm{cos}t$

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

$\mathcal{L}\left\{y\text{'}\text{'}-7y\text{'}+10y\right\}=\mathcal{L}\left\{9\mathrm{cos}t+7\mathrm{sin}t\right\}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}-7\mathcal{L}\left\{y\text{'}\right\}+10\mathcal{L}\left\{y\right\}=\frac{9s}{{s}^{2}+1}+\frac{7}{{s}^{2}+1}$

Solve for the transform as:

$\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-7\left[sY\left(s\right)-y\left(0\right)\right]+10Y\left(s\right)=\frac{9s+7}{{s}^{2}+1}\phantom{\rule{0ex}{0ex}}{s}^{2}Y\left(s\right)-5s+4-7\left[sY\left(s\right)-5\right]+10Y\left(s\right)\phantom{\rule{0ex}{0ex}}=\left({s}^{2}-7s+10\right)Y\left(s\right)=\frac{5{s}^{3}-39{s}^{2}+14s-32}{{s}^{2}+1}\phantom{\rule{0ex}{0ex}}Y\left(s\right)=\frac{5{s}^{3}-39{s}^{2}+14s-32}{\left({s}^{2}-7s+10\right)\left({s}^{2}+1\right)}$

Solve for the partial fraction as:

$\frac{5{s}^{3}-39{s}^{2}+14s-32}{\left({s}^{2}-7s+10\right)\left({s}^{2}+1\right)}=\frac{5{s}^{3}-39{s}^{2}+14s-32}{\left(s-5\right)\left(s-2\right)\left({s}^{2}+1\right)}\phantom{\rule{0ex}{0ex}}=\frac{A}{s-5}+\frac{B}{s-2}+\frac{Cs+D}{{s}^{2}+1}$

Solve further as:

role="math" localid="1664299653224" $5{s}^{3}-39{s}^{2}+14s-32=A\left({s}^{2}+1\right)\left(s-2\right)+B\left({s}^{2}+1\right)\left(s-5\right)+\phantom{\rule{0ex}{0ex}}\left(Cs+D\right)\left(s-5\right)\left(s-2\right)$

Using $s=5,2,0,1$ ,respectively, gives

$s=5:-312=78A⇒A=-4\phantom{\rule{0ex}{0ex}}s=2:-120=-15B⇒B=8\phantom{\rule{0ex}{0ex}}s=0:-32=-32+10D⇒D=0\phantom{\rule{0ex}{0ex}}s=1:-52=4C-56⇒C=1$

Therefore, the equation is:

$Y\left(s\right)=-\frac{4}{s-5}+\frac{8}{s-2}+\frac{s}{{s}^{2}+1}$

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

$y\left(t\right)={L}^{-1}\left\{-\frac{4}{s-5}+\frac{8}{s-2}+\frac{s}{{s}^{2}+1}\right\}\left(t\right)\phantom{\rule{0ex}{0ex}}=-4L\left\{\frac{1}{s-5}\right\}+8L\left\{\frac{1}{s-2}\right\}+C\left\{\frac{s}{{s}^{2}+1}\right\}\phantom{\rule{0ex}{0ex}}=-4{e}^{5t}+8{e}^{2t}+\mathrm{cos}t$

Therefore, the solution of the initial value problem is:

$y\left(t\right)=-4{e}^{5t}+8{e}^{2t}+\mathrm{cos}t$

Therefore, the initial value for $y\text{'}\text{'}-7y\text{'}+10y=9\mathrm{cos}t+7\mathrm{sin}t$ is $y\left(t\right)=-4{e}^{5t}+8{e}^{2t}+\mathrm{cos}t$