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10E

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$\mathbf{}\mathbf{10}\mathbf{.}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{t}\mathbf{-}\mathbf{8}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$

The Initial value for $y\text{'}\text{'}-4y=4t-8{e}^{-2t}isy\left(t\right)=-t-{e}^{-2t}+2t{e}^{-2t}+{e}^{2t}\phantom{\rule{0ex}{0ex}}$

See the step by step solution

Step 1: Define 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 as follows:

$\mathcal{L}\left\{y\text{'}\text{'}-4y\right\}=\mathcal{L}\left\{4t-8{e}^{-2t}\right\}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}-4\mathcal{L}\left\{y\right\}=\frac{4}{{s}^{2}}-\frac{8}{s+2}$

Solve the Laplace transform as:

$\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-4Y\left(s\right)=\frac{-8{s}^{2}+4s+8}{{s}^{2}\left(s+2\right)}\phantom{\rule{0ex}{0ex}}{s}^{2}Y\left(s\right)-5-4Y\left(s\right)=\frac{-8{s}^{2}+4s+8}{{s}^{2}\left(s+2\right)}\phantom{\rule{0ex}{0ex}}{s}^{2}Y\left(s\right)-4Y\left(s\right)=\frac{-8{s}^{2}+4s+8}{{s}^{2}\left(s+2\right)}+5\phantom{\rule{0ex}{0ex}}\left({s}^{2}-4\right)Y\left(s\right)=\frac{5{s}^{3}+2{s}^{2}+4s+8}{{s}^{2}\left(s+2\right)}\phantom{\rule{0ex}{0ex}}$

Solve further as:

$Y\left(s\right)=\frac{5{s}^{3}+2{s}^{2}+4s+8}{{s}^{2}{\left(s+2\right)}^{2}\left(s-2\right)}$

Using partial fractions solve as:

$\frac{5{s}^{3}+2{s}^{2}+4s+8}{{s}^{2}{\left(s+2\right)}^{2}\left(s-2\right)}=\frac{A}{{s}^{2}}+\frac{B}{s+2}+\frac{C}{{\left(s+2\right)}^{2}}+\frac{D}{s-2}$

Simplify the partial fractions as:

$5{s}^{3}+2{s}^{2}+4s+8=\left\{A{\left(s+2\right)}^{2}\left(s-2\right)+B{s}^{2}\left(s+2\right)\left(s-2\right)\phantom{\rule{0ex}{0ex}}+C{s}^{2}\left(s-2\right)+S-D{s}^{2}{\left(s+2\right)}^{2}\right\}$

Using s=0,2,1,-1 , respectively, gives

$s=0:8=-8A⇒A=-1\phantom{\rule{0ex}{0ex}}s=2:64=64D⇒D=1\phantom{\rule{0ex}{0ex}}s=1:19=18-3B-C⇒1=-3B-C\phantom{\rule{0ex}{0ex}}s=-1:1=4-3B-3C⇒3=3B+3C$

Find B and C from the system.

$\left\{\begin{array}{l}\begin{array}{l}3B-C=1\\ 3B+3C=3\end{array}\\ \end{array}\right\⇒\left\{\begin{array}{l}\begin{array}{l}C=2\\ B=-1\end{array}\\ \end{array}\right\$

$Therefore,Y\left(s\right)=-\frac{1}{{s}^{2}}-\frac{1}{s+2}+\frac{2}{{\left(s+2\right)}^{2}}+\frac{1}{s-2}\phantom{\rule{0ex}{0ex}}U\mathrm{sin}gtheinverseLaplacetransformweobtainthesolutionofgivendifferentialequation\phantom{\rule{0ex}{0ex}}y\left(t\right)={\mathcal{L}}^{-1}\left\{-\frac{1}{{s}^{2}}-\frac{1}{s+2}+\frac{2}{{\left(s+2\right)}^{2}}+\frac{1}{s-2}\right\}\left(t\right)\phantom{\rule{0ex}{0ex}}=-\mathcal{L}\left\{\frac{1}{{s}^{2}}\right\}-\mathcal{L}\left\{\frac{1}{s+2}\right\}+2\mathcal{L}\left\{\frac{1}{{\left(s+2\right)}^{2}}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{s-2}\right\}\phantom{\rule{0ex}{0ex}}=-t-{e}^{-2t}+2t{e}^{-2t}+{e}^{2t}\phantom{\rule{0ex}{0ex}}Therefore,theinitialvaluefory\text{'}\text{'}-4y=4t-8{e}^{-2t}isy\left(t\right)=-t-{e}^{-2t}+2t{e}^{-2t}+{e}^{2t}$