Find the transform of

f(t)={sin(x-vt),t>x/v0,t<x/v

Where xand vare constants.

Short Answer

Expert verified

The transform of Lf(t)=expv-vp2+v2 is .

Step by step solution

01

Given information

The given function isf(t)={sin(x-vt),t>x/v0,t<x/v

02

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)

03

Transform of the given function

The given expression is,

f(t)=sin(x-vt),t>x/v0,t<x/vf(t)=sin-vt-xv,t>x/v0,t<x/v

Use the expression

L28Lf(t)=e-pa.G(p)

Let

a=v=

And,

g(t-a)=sin-v(t-a)g(t)=sin(-vt)=-sin(vt)

Calculate the value of G (p).

G(p)=L[gt]=L[-sin(vt)]=-vp2+v2

Calculate the Laplace transform of the function.

Lf(t)=e-pa.G(p)=epv-vp2+v2

Thus, the Laplace transform ofsin(x-vt),t>x/vΩis localid="1659248824919" Lf(t)=epxv-vn2+v2.

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