Question

Prove the general formula L29.

Short answer

Answer

The function $L\left(e^{-at}g\left(f\right)\right)=G\left(p+a\right)$is proved

Step-by-step solution

Given information

The given function which has to prove is $L\left(f\right)={\int }_{0}f\left(t\right)e^{-pt}dt=F\left(p\right)$and function which has to prove is$L\left(e^{-at}g\left(f\right)\right)=G\left(p+a\right)$

Definition of Laplace Transformation

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Proof for the given function

It can be calculated as,

$$\begin{gathered} L\left(f\right)={\int }_0^{\infty }f\left(t\right)e^{-pt}dt \\ L\left(e^{-at}g\left(t\right)\right)={\int }_0^p{e}^{-at}{e}^{-pt}g\left(t\right)dt \\ L\left(e^{-at}g\left(t\right)\right)={\int }_0^p{e}^{-at}{e}^{-w-\mu t}g\left(t\right)dt \\ L\left(e^{-at}g\left(t\right)\right)={\int }_0^{\infty }{e}^{-\left(p+u\right)\square }g\left(t\right)dt \end{gathered}$$

Again, with use of

${\int }_0^{"}f\left(t\right)e^{-p}dt=F\left(p\right)$

$L\left(e^{-2}\left(t\right)\right)$Can be calculated as,

$$\begin{gathered} L\left(e^{-at}g\left(t\right)\right)={\int }_0{e}^{-\left(p+a\right)} \\ g\left(t\right)dt=G\left(p+a\right) \end{gathered}$$

Hence, $L\left(e^{-at}g\left(t\right)\right)=G\left(p+a\right)$.