Question

Findthe Laplace transform of$\mathbf{f}\mathbf{*}\mathbf{g}$. Do not evaluate the convolution integral before transforming. Where$\mathbf{L}\left\{t^2*{\mathrm{te}}^t\right\}$.

Short answer

The Laplace transformation is$L\left\{t^2*t{e}^t\right\}=\frac{2}{s^3{(s-1)}^2}$

Step-by-step solution

Perform the Laplace transformation;

The given function is$L\left\{t^2*t{e}^t\right\}$.

Use the result which is given below:

$\begin{array}{c}L\left\{f*g\right\}=L\left\{f\right\}L\left\{g\right\}\\ =F\left(s\right)G\left(s\right)\end{array}$

Using the formula to solve further;

Where f and g be piecewise continuous then,

$\begin{array}{c}L\left\{t^2*t{e}^t\right\}=L\left\{t^2\right\}.L\left\{t{e}^t\right\};\left[L\left\{t^n\right\}=\frac{n!}{s^{n+1}},L\left\{e^{at}f\left(t\right)\right\}={\left.F\left(s\right)\right|}_{s\to s-a}\right]\\ ={\left.\frac{2!}{s^3}.\frac{1}{s^2}\right|}_{s\to s-1}\\ =\frac{2!}{s^3}.\frac{1}{{\left(s-1\right)}^2}\\ =\frac{2}{s^3{\left(s-1\right)}^2}\end{array}$

The Laplace transformation is $L\left\{t^2*t{e}^t\right\}=\frac{2}{s^3{(s-1)}^2}$