Prove Theorem 7, page 368, on the linearity of the inverse transform. [Hint: Show that the right-hand side of equation (3) is a continuous function on $[0,\infty )$ whose Laplace transform is$\left[F_1\left(s\right)+F_2\left(s\right)\right]$

Given a function F(s) ,if there is a function f(t) that is continuous on $[0,\infty )$

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

Non-repeated Linear Factors

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

$Q\left(s\right)=\left(s-r_1\right)\left(s-r_2\right)...\left(s-r_n\right)$

where the role="math" localid="1664370001307" $r_i$'s are all distinct real numbers, then the partial fraction expansion has the form

$\frac{P\left(s\right)}{Q\left(s\right)}=\frac{A_1}{s-r_1}+\frac{A_2}{s-r_2}+...+\frac{A_n}{s-r_n}$

where the $A_i$'s are real numbers. There are various ways of determining the constants $A_1...A_n$. In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If s-r is a factor of Q(s) and ${\left(s-r\right)}^m$ is the highest power of s-r that divides Q(s), 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

$\frac{A_1}{s-r}+\frac{A_2}{{\left(s-r_1\right)}^2}+...+\frac{A_m}{{\left(s-r\right)}^m}$

where the$A_i$'s are real numbers.

Quadratic Factors

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 ${\left(s-\alpha \right)}^2+\beta$ that divides Q(s) , then the portion of the partial fraction expansion that corresponds to $$" width="9" height="19" role="math">s-α2+β2 is

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

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

$\frac{A_1\left(s-\alpha \right)+\beta B_1}{{\left(S-\alpha \right)}^2+{\beta }^2}+\frac{A_2\left(s-\alpha \right)+\beta B_2}{{\left[{\left(S-\alpha \right)}^2+{\beta }^2\right]}^2}+...+\frac{A_M\left(s-\alpha \right)+\beta B_M}{{\left[{\left(S-\alpha \right)}^2+{\beta }^2\right]}^m}$

Since ${\mathcal{L}}^{-1}\left\{F_1\right\}$and ${\mathcal{L}}^{-1}\left\{F_2\right\}$are both continuous functions on $[0,\infty )$we have that their sum also is continuous function on $[0,\infty )$

Note that

$$\begin{gathered} \mathcal{L}\left\{{\mathcal{L}}^{-1}\left\{F_1\right\}+{\mathcal{L}}^{-1}\left\{F_2\right\}\right\}\left(s\right)=\mathcal{L}\left\{{\mathcal{L}}^{-1}\left\{F_1\right\}\right\}\left(s\right)+\mathcal{L}\left\{{\mathcal{L}}^{-1}\left\{F_2\right\}\right\}\left(s\right) \\ {F}_1\left(s\right)+F_2\left(s\right) \end{gathered}$$

and

$c\mathcal{L}\left\{{\mathcal{L}}^{-1}\left(F\right)\right\}\left\{s\right\}=cF\left(s\right)$

from where it follows that

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{F_1+F_2\right\}={\mathcal{L}}^{-1}\left\{F_1\right\}+{\mathcal{L}}^{-1}\left\{F_2\right\}{\mathcal{L}}^{-1}\left\{cF\right\} \\ =c{\mathcal{L}}^{-1}\left\{F\right\} \end{gathered}$$

It is proved that:

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{F_1+F_2\right\}={\mathcal{L}}^{-1}\left\{F_1\right\}+{\mathcal{L}}^{-1}\left\{F_2\right\} \\ {\mathcal{L}}^{-1}\left\{cF\right\}=c{\mathcal{L}}^{-1}F \end{gathered}$$