Question

t e^2t. sin 3t find Laplace transform​

Answer 1

1. L(sin 3t)= 3/ (s^2 + 9 ) as L(sinat = a / (s^2 + a^2 )
2. L(e^2t sin 3t) = 3/ [(s-2)^2 + 9]
3. L( te^2t sin 3t) = -d/ds [3/ [(s-2)^2 + 9]
4. = {{[(s-2)^2 + 9] × 0} - [3(2s - 4)] } / [(s-2)^2 + 9]^2

(On differentiating)

5. =6(s-2)/ ([(s-2)^2 + 9]^2 ( ANS )

Answer 2

Given,

equation : $te^{2t}.sin3t$                                  

To Find,

the Laplace transform of the given equation $te^{2t}.sin3t$

Solution,

applying Laplace to the given equation,

$L(sin3t) = \frac{3}{(s^{2}+9) }$                    [using $L(sinat)=\frac{a}{(s^{2}+a^{2}) }$ ]

$L(e^{2t}sin3t)=\frac{3}{[(s-2)^{2} +9]} \\\\L(te^{2t}sin3t)= -\frac{d}{ds}[\frac{3}{[(s-2)^{2}+9] }]$

                   =$[[[(s-2)^{2}+9]$ X 0$]-[\frac{3(2s-4)}{[(s-2)^{2}+9 )^{2}] }]$

on differentiating

 $L(te^{2t}.sin3t)= \frac{6(s-2)}{[(s-2)^{2}+9]^{2} }$

Hence the Laplace transformation of the equation $te^{2t}.sin3t$ is $\frac{6(s-2)}{[(s-2)^{2}+9]^{2} }$ .