Question

a)With a slight changes in notation the transform in(6) is the same as $\mathit{L}\left\{f\text{'}\left(t\right)\right\}\mathbf{=}\mathit{s}\mathit{L}\left\{f\left(t\right)-f\left(0\right)\right\}$with $\mathit{f}\left(t\right)\mathbf{=}\mathit{a}\mathit{t}{\mathit{e}}^{\mathbf{a}\mathbf{t}}$,discuss how this result in conjunction with ©of Theorem can 7.11 can be used to evaluate localid="1663977247068" $\mathit{L}\mathbf{\left\{}{\mathbf{te}}^{\mathbf{at}}\mathbf{\right\}}$

b)proceed as in part (a).but this time discuss how to use (7) with $\mathit{f}\left(t\right)\mathbf{=}\mathit{t}\mathit{s}\mathit{i}\mathit{n}\mathit{k}\mathit{t}$ in conjunction with(d) and(e)of theorem to evaluate $\mathit{L}\left\{tsinkt\right\}\mathbf{.}$

Short answer

$L\left\{f\text{'}\left(t\right)\right\}=L\left\{ate+e^{at}\right\}$Laplace transform of both side of the above equation

a)$L\left\{t{e}^{at}\right\}\frac{1}{s-a}$

Differentiate the given equation with respect to t,

Take Laplace transform of both side of the above equation. So, (b)

$f\text{'}\left(t\right)=2k\cos kt+k^{2}t\sin kt$

b)$L\left\{tsinkt\right\}=\frac{2ks}{{\left(s^2+k^2\right)}^2}$

Step-by-step solution

Step1: definition of Laplace transformation.

A transformation of a function f(x) into the function is particularly useful for reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Differentiate the given equation . Laplace transform of both side of the above equation

$f\text{'}\left(t\right)=at{e}^{at}+e^{at}$

Take Laplace transform of both side of the above equation

$$\begin{gathered} L\left\{f\text{'}\left(t\right)\right\}=L\left\{ate+e^{at}\right\} \\ \begin{array}{l}aL\left\{t{e}^{at}\right\}+L\left\{e^{at}\right\}\\ aL\left\{t{e}^{at}\right\}+\frac{1}{s-a}\left(L\left\{e^{at}\right\}=\frac{1}{s-a}\right)\\ sL\left\{f\left(t\right)\right\}=aL\left\{t{e}^{at}\right\}+\frac{1}{s-a}\\ |\left(f\left(t\right)=t{e}^{at},f\left(0\right)=0\right)\\ sL\left\{f\left(t\right)\right\}-f\left(0\right)=aL\left\{t{e}^{at}\right\}+\frac{1}{s-a}\\ \left(s-a\right)L\left\{t{e}^{at}\right\}\frac{1}{s-a}\end{array} \\ L\left\{t{e}^{at}\right\}\frac{1}{s-a} \end{gathered}$$

 Differentiate the given equation.

Differentiate the given equation with respect to t,

Andagaindifferentiate withrespect to t.

$f\text{'}\left(t\right)=\sin kt+tk\cos kt$

Take Laplace transform of both side of the above equation. So, (b)

$f\text{'}\left(t\right)=2k\cos kt+k^{2}t\sin kt$

$\begin{array}{c}L\left\{f\text{'}\text{'}\left(t\right)\right\}=L\left\{f\text{'}\left(t\right)=2k\cos kt+k^{2}t\sin kt\right\}\\ =2kL\left\{\cos kt\right\}-k^{2}L\left\{t\sin kt\right\}\\ =2k\frac{s}{s^2+k^2}-k^{2}L\left\{t\sin kt\right\}\\ =2k\left(\frac{s}{s^2+b^2}=L\left\{\cos kt\right\}\right)\\ \end{array}$

So,

$\begin{array}{c}L\left\{f\text{'}\text{'}\left(t\right)\right\}=2k\frac{s}{s^2+k^2}-k^{2}L\left\{tsinkt\right\}\\ s^2\left\{f\left(t\right)\right\}-sf\left(0\right)-f\text{'}\left(0\right)=2k\frac{s}{s^2+k^2}-k^{2}L\left\{tsinkt\right\}\\ \left(s^2+k^2\right)L\left\{tsinkt\right\}=\frac{2ks}{s^2+k^2}\\ L\left\{tsinkt\right\}=\frac{2ks}{{\left(s^2+k^2\right)}^2}\end{array}$

Therefore,

a)$L\left\{t{e}^{at}\right\}\frac{1}{s-a}$

b)$L\left\{tsinkt\right\}=\frac{2ks}{{\left(s^2+k^2\right)}^2}$$L\left\{tsinkt\right\}=\frac{2ks}{{\left(s^2+k^2\right)}^2}$