Question

Use L29 and L11 to obtain $\mathit{L}\left(t{e}^{-at}sinbt\right)$which is not in the table.

Short answer

Answer

The solution is $L\left\{t{e}^{-at}\sin br\right\}=\frac{2b\left(p+a\right)}{{\left({\left(w+a\right)}^2+b^2\right)}^2}$

Step-by-step solution

Given information

A function as $f\left(t\right)=t{e}^{at}\sin bt$

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

Formula used for calculations is :

$$\begin{gathered} L\left(29\right):L\left\{e^{-B_g\left(t\right)}\right\}-G\left(p+a\right) \\ L\left(11\right):L\left\{t\sin at\right\}=\frac{24p}{{\left(p^2+{\omega }^2\right)}^2} \end{gathered}$$

Proof for given function

From (L29) $L\left\{e^{-dt}g\left(t\right)\right\}=G\left(p+a\right)$

Since from (L11)$L\left\{t\sin bt\right\}=\frac{2h}{{\left(p^2+b^2\right)}^2}$

Consider $g\left(t\right)=t\sin bt$and by use of (L29),

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

$L\left\{t{e}^{at}\sin bt\right\}$Can be calculated by replace p by $\left(p+a\right)$in equation(1) as,

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

Hence $L\left\{t{e}^{-at}\sin bt\right\}=\frac{2b\left(p+a\right)}{{\left({\left(p+a\right)}^2+b^2\right)}^2}$