## 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}

As

\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.

## 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!