Question

A) Use the definition of Laplace transform to find L{f }. (Do the integrals.) For what values of s is L{f } defined?f(t) = (2t+1)

Answer 1

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = $\int\limits^\infty_0 {f(t)e^{-st} } \, dt$

          =  $\int\limits^\infty_0[2t +1] e^{-st} dt$

          = $2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt$

         = 2 L(t) + L(1)

L(1) = $\int\limits^\infty_0e^{-st} dt$

     = (-1/s) ( 0 -1 )

     = 1/s , ( s >  0)

2L ( t ) = $2\int\limits^\infty_0 te^{-st}$

        =  $2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]$

        =  2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

          = 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

brainly.com/question/14487937

#SPJ4