Step 1

Given

Convolution theorem.

Step 2

The convolution theorem states that,

If \({L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={F}{\left({s}\right)}\ \text{ and }\ {L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}={G}{\left({s}\right)}\) then

\({L}{\left\lbrace f{{\left({t}\right)}}\ast g{{\left({t}\right)}}\right\rbrace}={F}{\left({s}\right)}{G}{\left({s}\right)}\ \text{ where }\ f{{\left({t}\right)}}\ast g{{\left({t}\right)}}={\int_{{0}}^{{t}}} f{{\left({r}\right)}} g{{\left({t}-{r}\right)}}{d}{r}\)

Simply the convolution theorem means if we have two functions then taking their convolution and then Laplace is the same as taking the Laplace first of the two functions separately and then multiplying the two Laplace transforms.

Convolution theorem applied to the solution of differential equation.

Convolution theorem is useful in differentiation and integration of Laplace transforms.

Given

Convolution theorem.

Step 2

The convolution theorem states that,

If \({L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={F}{\left({s}\right)}\ \text{ and }\ {L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}={G}{\left({s}\right)}\) then

\({L}{\left\lbrace f{{\left({t}\right)}}\ast g{{\left({t}\right)}}\right\rbrace}={F}{\left({s}\right)}{G}{\left({s}\right)}\ \text{ where }\ f{{\left({t}\right)}}\ast g{{\left({t}\right)}}={\int_{{0}}^{{t}}} f{{\left({r}\right)}} g{{\left({t}-{r}\right)}}{d}{r}\)

Simply the convolution theorem means if we have two functions then taking their convolution and then Laplace is the same as taking the Laplace first of the two functions separately and then multiplying the two Laplace transforms.

Convolution theorem applied to the solution of differential equation.

Convolution theorem is useful in differentiation and integration of Laplace transforms.