themobius6s

2022-09-08

If Fi(t) and fi(s) are Laplace pairs, show that:

$$\mathcal{L}\{\sum _{i=1}^{n}{F}_{i}(t)\}=\sum _{i=1}^{n}\mathcal{L}\left\{{F}_{i}\right\}=\sum _{i=1}^{n}{f}_{i}(s)$$

$$\mathcal{L}\{\sum _{i=1}^{n}{F}_{i}(t)\}=\sum _{i=1}^{n}\mathcal{L}\left\{{F}_{i}\right\}=\sum _{i=1}^{n}{f}_{i}(s)$$

Ufumanaxi

Beginner2022-09-09Added 5 answers

It comes from the linearity of the integral

$$\begin{array}{rl}\mathcal{L}\{\sum _{i=1}^{n}{F}_{i}(t)\}& ={\int}_{0}^{\mathrm{\infty}}(\sum _{i=1}^{n}{F}_{i}(t)){\mathrm{e}}^{st}\mathrm{d}t\\ & ={\int}_{0}^{\mathrm{\infty}}({F}_{1}(t)+\cdots +{F}_{n}(t)){\mathrm{e}}^{st}\mathrm{d}t\\ & ={\int}_{0}^{\mathrm{\infty}}{F}_{1}(t){\mathrm{e}}^{st}\mathrm{d}t+\cdots +{\int}_{0}^{\mathrm{\infty}}{F}_{n}(t){\mathrm{e}}^{st}\mathrm{d}t\\ & =\mathcal{L}\{{F}_{1}(t)\}+\cdots +\mathcal{L}\{{F}_{n}(t)\}\\ & =\sum _{i=1}^{n}\mathcal{L}\{{F}_{i}(t)\}\\ & =\sum _{i=1}^{n}{f}_{i}(s)\end{array}$$

$$\begin{array}{rl}\mathcal{L}\{\sum _{i=1}^{n}{F}_{i}(t)\}& ={\int}_{0}^{\mathrm{\infty}}(\sum _{i=1}^{n}{F}_{i}(t)){\mathrm{e}}^{st}\mathrm{d}t\\ & ={\int}_{0}^{\mathrm{\infty}}({F}_{1}(t)+\cdots +{F}_{n}(t)){\mathrm{e}}^{st}\mathrm{d}t\\ & ={\int}_{0}^{\mathrm{\infty}}{F}_{1}(t){\mathrm{e}}^{st}\mathrm{d}t+\cdots +{\int}_{0}^{\mathrm{\infty}}{F}_{n}(t){\mathrm{e}}^{st}\mathrm{d}t\\ & =\mathcal{L}\{{F}_{1}(t)\}+\cdots +\mathcal{L}\{{F}_{n}(t)\}\\ & =\sum _{i=1}^{n}\mathcal{L}\{{F}_{i}(t)\}\\ & =\sum _{i=1}^{n}{f}_{i}(s)\end{array}$$

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

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