Question

Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform of

$F\left(s\right)=\frac{3s^2-16s+5}{(s+1)(s-3)(s-2)}$

Short answer

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

Step-by-step solution

Define Inverse Laplace transform

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

$[0,\infty )$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 $(s-r{)}^m$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 $(s-r{)}^m$ is

$\frac{A_1}{s-r}+\frac{A_2}{{(s-r)}^2}+\cdots +\frac{A_m}{{(s-r)}^m},$

where the $A_i$'s are real numbers.

Quadratic Factors

If $(s-\alpha {)}^2+{\beta }^2$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 $(s-\alpha {)}^2+{\beta }^2$ that divides $Q\left(s\right)$, then the portion of the partial fraction expansion that corresponds to$(s-\alpha {)}^2+{\beta }^2$ is

$\frac{C_{1}s+D_1}{{(s-\alpha )}^2+{\beta }^2}+\frac{C_{2}s+D_2}{{\left[{(s-\alpha )}^2+{\beta }^2\right]}^2}+\cdots +\frac{C_{m}s+D_m}{{\left[{(s-\alpha )}^2+{\beta }^2\right]}^m}.$

it is more convenient to express $C_{i}s+D_i$in the form $A_i(s-\alpha )+\beta B_i$when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{A_1(s-\alpha )+\mathbf{\beta }{B}_1}{{(s-\alpha )}^2+{\mathbf{\beta }}^2}+\frac{A_2(s-\alpha )+\mathbf{\beta }{B}_2}{{\left[{(s-\alpha )}^2+{\mathbf{\beta }}^2\right]}^2}+\cdots +\frac{{\mathbf{A}}_m(s-\alpha )+\mathbf{\beta }{B}_m}{{\left[{(s-\alpha )}^2+{\mathbf{\beta }}^2\right]}^m}.$

Find the factor of the denominator

There we have

$$\begin{gathered} P\left(s\right)=3s^2-16s+5 \\ Q\left(s\right)=(s+1)(s-3)(s-2) \end{gathered}$$

SO

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

Now we have

$$\begin{gathered} Q\left(-1\right)=12P\left(-1\right)=24 \\ {Q}^{\text{'}}\left(3\right)=4P\left(3\right)=-16 \\ {Q}^{\text{'}}\left(2\right)=-3P\left(2\right)=-15 \end{gathered}$$

Using the equation from Problem 40 we get

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

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