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Expert-verified Found in: Page 1 ### 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 # Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform of${F}{\left(}{s}{\right)}{=}\frac{3{s}^{2}-16s+5}{\left(s+1\right)\left(s-3\right)\left(s-2\right)}$

$2{e}^{-t}-4{e}^{3t}+5{e}^{2t}$

See the step by step solution

## Step 1: Define Inverse Laplace transform

Given a function $F\left(s\right)$ , if there is a function $f\left(t\right)$ that is continuous on

$\left[0,\infty \right)$ and satisfies$\mathcal{L}\left\{f\right\}=F$,then we say that $f\left(t\right)$ is the inverse Laplace transform of $F\left(s\right)$and employ the notation $f={\mathcal{L}}^{-1}\left\{F\right\}$

Non-repeated Linear Factors

If $Q\left(s\right)$ can be factored into a product of distinct linear factors,

$Q\left(s\right)=\left(s-{r}_{1}\right)\left(s-{r}_{2}\right)\cdots \left(s-{r}_{n}\right),$

where the ${r}_{i}$ 's are all distinct real numbers, then the partial fraction expansion has the form

$\frac{P\left(s\right)}{Q\left(s\right)}=\frac{{A}_{1}}{s-{r}_{1}}+\frac{{A}_{2}}{s-{r}_{2}}+\cdots +\frac{{A}_{n}}{s-{r}_{n}},$

where the ${A}_{i}$ 's are real numbers. There are various ways of determining the constants ${A}_{1},\dots ,{A}_{n}$. In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If $s-r$ is a factor of $Q\left(s\right)$ and $\left(s-r{\right)}^{m}$ is the highest power of $s-r$ that divides $Q\left(s\right)$, then the portion of the partial fraction expansion of $P\left(s\right)/Q\left(s\right)$ that corresponds to the term $\left(s-r{\right)}^{m}$ is

$\frac{{A}_{1}}{s-r}+\frac{{A}_{2}}{{\left(s-r\right)}^{2}}+\cdots +\frac{{A}_{m}}{{\left(s-r\right)}^{m}},$

where the ${A}_{i}$'s are real numbers.

If $\left(s-\alpha {\right)}^{2}+{\beta }^{2}$ is a quadratic factor of $Q\left(s\right)$ that cannot be reduced to linear factors with real coefficients and $m$ is the highest power of $\left(s-\alpha {\right)}^{2}+{\beta }^{2}$ that divides $Q\left(s\right)$ , then the portion of the partial fraction expansion that corresponds to $\left(s-\alpha {\right)}^{2}+{\beta }^{2}$ is

$\frac{{C}_{1}s+{D}_{1}}{{\left(s-\alpha \right)}^{2}+{\beta }^{2}}+\frac{{C}_{2}s+{D}_{2}}{{\left[{\left(s-\alpha \right)}^{2}+{\beta }^{2}\right]}^{2}}+\cdots +\frac{{C}_{m}s+{D}_{m}}{{\left[{\left(s-\alpha \right)}^{2}+{\beta }^{2}\right]}^{m}}.$

it is more convenient to express ${C}_{i}s+{D}_{i}$ in the form ${A}_{i}\left(s-\alpha \right)+\beta {B}_{i}$when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{{A}_{1}\left(s-\alpha \right)+\mathbf{\beta }{B}_{1}}{{\left(s-\alpha \right)}^{2}+{\mathbf{\beta }}^{2}}+\frac{{A}_{2}\left(s-\alpha \right)+\mathbf{\beta }{B}_{2}}{{\left[{\left(s-\alpha \right)}^{2}+{\mathbf{\beta }}^{2}\right]}^{2}}+\cdots +\frac{{\mathbf{A}}_{m}\left(s-\alpha \right)+\mathbf{\beta }{B}_{m}}{{\left[{\left(s-\alpha \right)}^{2}+{\mathbf{\beta }}^{2}\right]}^{m}}.$

## Step 2: Find the factor of the denominator

There we have

$P\left(s\right)=3{s}^{2}-16s+5\phantom{\rule{0ex}{0ex}}Q\left(s\right)=\left(s+1\right)\left(s-3\right)\left(s-2\right)$

SO

${Q}^{\text{'}}\left(s\right)=\left(s-3\right)\left(s-2\right)+\left(s+1\right)\left(s-2\right)+\left(s+1\right)\left(s-3\right)$

Now we have

$Q\left(-1\right)=12P\left(-1\right)=24\phantom{\rule{0ex}{0ex}}{Q}^{\text{'}}\left(3\right)=4P\left(3\right)=-16\phantom{\rule{0ex}{0ex}}{Q}^{\text{'}}\left(2\right)=-3P\left(2\right)=-15$

Using the equation from Problem 40 we get

${\mathcal{L}}^{-1}\left\{\frac{3{s}^{2}-16s+5}{\left(s+1\right)\left(s-3\right)\left(s-2\right)}\right\}\left(t\right)=\frac{P\left(-1\right)}{{Q}^{\text{'}}\left(-1\right)}{e}^{-t}+\frac{P\left(3\right)}{{Q}^{\text{'}}\left(3\right)}{e}^{3t}+\frac{P\left(2\right)}{{Q}^{\text{'}}\left(2\right)}{e}^{2t}\phantom{\rule{0ex}{0ex}}=\frac{24}{12}{e}^{-t}+\frac{-16}{4}{e}^{3t}+\frac{-15}{-3}{e}^{2t}\phantom{\rule{0ex}{0ex}}=2{e}^{-t}-4{e}^{3t}+5{e}^{2t}$

Thus, ${\mathcal{L}}^{-1}\left\{\frac{3{s}^{2}-16s+5}{\left(s+1\right)\left(s-3\right)\left(s-2\right)}\right\}\left(t\right)=2{e}^{-t}-4{e}^{3t}+5{e}^{2t}$ ### Want to see more solutions like these? 