Question

Obtain $\mathit{L}\left(t{e}^{-at}cosbt\right)$

Short answer

Answer

The solution is $L\left\{t{e}^{-at}\cos bt\right\}-\frac{{\left(p+a\right)}^2-b^2}{{\left({\left(\phi +a\right)}^2+b^2\right)}^2}$

Step-by-step solution

Given information

The given function is $f\left(t\right)=t{e}^{-at}\cos bt$and to prove is$L\left(t{e}^{at}\cos bt\right)$

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)

Properties used for given function

$$\begin{gathered} L\left(29\right):L\left\{e^{-at}g\left(t\right)\right\}=G\left(p+a\right) \\ L\left(12\right):L\left\{t\cos at\right\}=\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2} \end{gathered}$$

Proof for given function

From L29

$L\left\{e^{-a{i}_g}g\left(t\right)\right\}=G\left(p+a\right)$

Since from L11

$L\left\{e^{-a{i}_g}g\left(t\right)\right\}=G\left(p+a\right)$

Consider $g\left(t\right)=t\cos bt$and by use of (L29), L $\left\{e^{-at}g\left(t\right)\right\}=G\left(p+a\right);L\left\{t{e}^{-w}\cos bt\right\}$can be calculated by replace $pby\left(p+a\right)$in equation as,

$$\begin{gathered} L\left\{t\cos bt\right\}=\frac{p^2-b^2}{{\left(p^2+b^2\right)}^2} \\ L\left\{t{e}^{-at}\cos bt\right\}=\frac{{\left(p+a\right)}^2-b^2}{{\left({\left(p+a\right)}^2+b^2\right)}^2} \end{gathered}$$

Hence, $L\left\{t_e^{-at}\cos bt\right\}=\frac{{\left(p+a\right)}^2-b^2}{{\left({\left(p+a\right)}^2\right)}^2}$