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9E

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 ${\mathbf{1}}{\mathbf{-}}{\mathbf{14}}$ , solve the given initial value problem using the method of Laplace transforms. ${\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{5}}{\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{6}}{\mathbf{z}}{\mathbf{=}}{\mathbf{21}}{{\mathbf{e}}}^{\mathbf{t}\mathbf{-}\mathbf{1}}{\mathbf{,}}{}{\mathbf{z}}\left(\mathbf{1}\right){\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{,}}{}{}{}{\mathbf{z}}{\mathbf{\text{'}}}\left(\mathbf{1}\right){\mathbf{=}}{\mathbf{9}}$

The Initial value for $z\text{'}\text{'}+5z\text{'}-6z=21{e}^{t-1}$ is $z\left(t\right)=3\left(t-1\right){e}^{t-1}-{e}^{-6\left(t-1\right)}$

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

Shift the initial conditions to $t=0$ by defining a new function:

$y\left(t\right)=z\left(t+1\right)\phantom{\rule{0ex}{0ex}}y\text{'}\left(t\right)=z\text{'}\left(t+1\right)\phantom{\rule{0ex}{0ex}}y\text{'}\text{'}\left(t\right)=z\text{'}\text{'}\left(t+1\right)$

Replace $t$ by $t+1$ in the condition

$z\text{'}\text{'}\left(t\right)+5z\text{'}\left(t\right)-6z\left(t\right)=21{e}^{t-1}\phantom{\rule{0ex}{0ex}}⇒z\text{'}\text{'}\left(t+1\right)+5z\text{'}\left(t+1\right)-6z\left(t+1\right)=21{e}^{\left(t+1\right)-1}$

$⇒z\text{'}\text{'}\left(t+1\right)+5z\text{'}\left(t+1\right)-6z\left(t+1\right)=21{e}^{t}$

Substitute $y\left(t\right)=z\left(t+1\right)$

Solve for initial condition.

$y\left(0\right)=z\left(0+1\right)\phantom{\rule{0ex}{0ex}}=z\left(1\right)\phantom{\rule{0ex}{0ex}}=-1$

Solve for differentiation of initial condition.

$y\text{'}\left(0\right)=z\text{'}\left(0+1\right)\phantom{\rule{0ex}{0ex}}=z\text{'}\left(1\right)\phantom{\rule{0ex}{0ex}}=9$

Simplify the equation as:

$y\text{'}\text{'}\left(t\right)+5y\text{'}\left(t\right)-6y\left(t\right)=21{e}^{t}\phantom{\rule{0ex}{0ex}}y\left(0\right)=-1\phantom{\rule{0ex}{0ex}}y\text{'}\left(0\right)=9$

Define $\mathcal{L}\left\{y\right\}\left(s\right)=Y\left(s\right)$

sing the properties listed below, take the Laplace transform of the equation

$\mathcal{L}\left\{y\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{y\right\}\left(s\right)-y\left(0\right)\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}\left(s\right)={s}^{2}\mathcal{L}\left\{y\right\}\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)$

$\mathcal{L}\left\{{e}^{at}\right\}\left(s\right)=\frac{1}{s-a}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}+5L\left\{y\text{'}\right\}-6L\left\{y\right\}=21L\left\{{e}^{t}\right\}$

Substitute the properties into the equation.

$\left[{s}^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)\right]+5\left[sY-y\left(0\right)\right]-6Y=\frac{21}{s-1}\phantom{\rule{0ex}{0ex}}$

Substitute the initial conditions:

$y\left(0\right)=-1andy\text{'}\left(0\right)=9$

$\left({s}^{2}Y+s-9\right)+5\left(sY+1\right)-6Y=\frac{21}{s-1}$

Distribute and simplify:

$⇒{s}^{2}Y+5sY-6Y+s-4=\frac{21}{s-1}$

Isolate the Y variable.

$Y\left({s}^{2}+5s-6\right)=\frac{21}{s-1}-s+4\phantom{\rule{0ex}{0ex}}⇒Y\left({s}^{2}+5s-6\right)=\frac{21+\left(-s+4\right)\left(s-1\right)}{s-1}\phantom{\rule{0ex}{0ex}}⇒Y=\frac{-{s}^{2}+5s+17}{\left(s-1\right)\left({s}^{2}+5s-6\right)}$

Find the partial fraction expansion

Because $s-1$ is a repeated factor of ${\left(s-1\right)}^{2}\left(s+6\right)$ ,we include $\left(s-1\right)$ and ${\left(s-1\right)}^{2}$

$Y=\frac{-{s}^{2}+5s+17}{\left(s-1\right)\left(s-1\right)\left(s+6\right)}\phantom{\rule{0ex}{0ex}}$

$⇒\frac{-{s}^{2}+5s+17}{{\left(s-1\right)}^{2}\left(s+6\right)}=\frac{A}{s-1}+\frac{B}{{\left(s-1\right)}^{2}}+\frac{C}{s+6}$

Combine the fractions to equate the numerators.

$-{s}^{2}+5s+17=A\left(s-1\right)\left(s+6\right)+B\left(s+6\right)+C{\left(s-1\right)}^{2}$

Solve for variables by setting values of S

$s=1⇒21=A·0+7B+C·0=7B⇒B=3\phantom{\rule{0ex}{0ex}}s=-6⇒-49=A·0+3·0+49C⇒C=-1$

$s=0⇒17=-6A+3·6-1·1⇒A=0$

Substitute the values A,B,C of into partial fraction expansion

Using the properties listed below take the inverse Laplace transform to obtain the solution $y\left(t\right)$

${c}^{-1}\left\{\frac{1}{s-a}\right\}={e}^{at}\phantom{\rule{0ex}{0ex}}{c}^{-1}\left\{\frac{n!}{{\left(s-a\right)}^{n+1}}\right\}={e}^{at}{t}^{n}$

Solve for the solution of differential equation as:

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

Since , $y\left(t-1\right)=z\left(t\right)$, , replace $t$ by $t-1$

$y\left(t-1\right)=z\left(t\right)\phantom{\rule{0ex}{0ex}}=3\left(t-1\right){e}^{t-1}-{e}^{-6\left(t-1\right)}$

Therefore, the Initial value for $z\text{'}\text{'}+5z\text{'}-6z=21{e}^{t-1}$is $z\left(t\right)=3\left(t-1\right){e}^{t-1}-{e}^{-6\left(t-1\right)}$