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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

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, ${\mathcal{L}}^{-1}\left\{\frac{3{s}^{2}-16s+5}{\left(s+1\right)\left(s-3\right)\left(s-2\right)}\right\}\left(t\right)=2{e}^{-t}-4{e}^{3t}+5{e}^{2t}$

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