Linearity of Laplace transform - laplace-transform.com

1. Intuitive introduction

In elementary calculus, we learn the integral is a linear application:

$\displaystyle\int_{a}^{b}[\alpha\cdot f(x)+g(x)]dx=\alpha\cdot\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$

where $\alpha\in\mathbb{R}$ .

Now, let $f(t)$ and $g(t)$ be two functions for $t\geq0$ . Suppose the Laplace Transform of each of them can be evaluated, i.e., the integrals below converge for some $s$ :

$\displaystyle \mathscr{L}\{f(t)\}=F(s)=\int_{0}^{\infty}e^{-st}f(t)dt$

$\displaystyle \mathscr{L}\{g(t)\}=G(s)=\int_{0}^{\infty}e^{-st}g(t)dt$

Hence, the integral linearity works here too:

$\begin{aligned} \int_{0}^{\infty} e^{-st}[\alpha\cdot f(t)+g(t)]dt & = \int_{0}^{\infty}(e^{-st}\cdot\alpha\cdot f(t)+e^{-st}\cdot g(t))dt \\& = \alpha\cdot\int_{0}^{\infty}e^{-st}f(t)dt+\int_{0}^{\infty}e^{-st}g(t)dt\end{aligned}$

$\displaystyle\mathscr{L}\{\alpha\cdot f(t)+g(t)\}=\int_{0}^{\infty} e^{-st}[\alpha\cdot f(t)+g(t)]dt$

we have the following theorem.

2. Linearity Theorem

3. Illustrative example

we can use the Laplace linearity:

$\mathscr{L}\{1+4e^{3t}\}=\mathscr{L}\{1\}+4\mathscr{L}\{e^{3t}\}$

The Laplace Transform of 1 is:

$\displaystyle\frac{1}{s}$

and as (check proof):

$\displaystyle\mathscr{L}\{e^{at}\}=\frac{1}{s-a}$

so for $a=3$ , we have:

$\displaystyle\mathscr{L}\{e^{3t}\}=\frac{1}{s-3}$

Thus,

$\displaystyle\mathscr{L}\{1+4e^{3t}\}=\frac{1}{s}+\frac{4}{s-3}$

We reached the end of this short lesson about Laplace transform linearity property. For more properties of Laplace transform, check this post!