Question

Find the Laplace transform of the following function
\[g(t)=t^2\sinh(3t)\]

Answer 1

\[\begin{align*}
\\\text{(b)}&&g(t)=t^2\sinh(3t)\\
\\&&G(p)=\mathcal L\{t^2\sinh(3t)\}\\
\\&&=\int\limits_0^\infty t^2\sinh(3t)e^{-pt}\text dt\\
\\&&=\int\limits_0^\infty t^2\left(\frac{e^{3t}-e^{-3t}}{2}\right)e^{-pt}\text dt\\
\\&&=\frac12\int\limits_0^\infty t^2\left(e^{3t}-e^{-3t}\right)e^{-pt}\text dt\\
\\&&=\frac12\left[\int\limits_0^\infty t^2e^{-(p-3)}\text dx-\int\limits_0^\infty t^2e^{-(p+3)}\text dt\right]\\
\\&&=\frac12\left[\frac{2!}{(p-3)^3}-\frac{2!}{(p+3)^3}\right]\\
\\&&=\frac{1}{(p-3)^3}-\frac{1}{(p+3)^3}\\
\\&&=\frac{(p+3)^3-(p-3)^3}{(p-3)^3(p+3)^3}\\
\\&&=\frac{(p+3)(p^2+6p+9)-(p-3)(p^2-6p-9)}{\left((p-3)(p+3)\right)^3}\\
\\&&=\frac{(p^3+6p^2+9p+3p^2+18p+27)-(p^3-6p^2-9p-3p^2+18p+27)}{\left(p^2-9\right)^3}\\
\\&&=\frac{18p^2+18p}{\left(p^2-9\right)^3}\\
\\&&=\frac{18p(p+1)}{\left(p^2-9\right)^3}\\
\end{align*}\]

Answer 2

or should i leave my solution as
\[G(p)=\frac{1}{(p-3)^3}-\frac{1}{(p+3)^3}\]

Answer 3

?

Answer 4

Don't know.. :)

Answer 5

Your work is cut off pretty badly

Answer 6

don't leave that answer the difference form, what simplification u have done was required, i think...

Answer 7

I can't see the rest of the steps @UnkleRhaukus how can i see that. ?

Answer 8

\(\begin{align*}
\\\text{(b)} &g(t)=t^2\sinh(3t)\\
\\ &G(p)=\mathcal L\{t^2\sinh(3t)\}\\
\\ &=\int\limits_0^\infty t^2\sinh(3t)e^{-pt}\text dt\\
\\ &=\int\limits_0^\infty t^2\left(\frac{e^{3t}-e^{-3t}}{2}\right)e^{-pt}\text dt\\
\\ &=\frac12\int\limits_0^\infty t^2\left(e^{3t}-e^{-3t}\right)e^{-pt}\text dt\\
\\ &=\frac12\left[\int\limits_0^\infty t^2e^{-(p-3)}\text dx-\int\limits_0^\infty t^2e^{-(p+3)}\text dt\right]\\
\\ &=\frac12\left[\frac{2!}{(p-3)^3}-\frac{2!}{(p+3)^3}\right]\\
\\ &=\frac{1}{(p-3)^3}-\frac{1}{(p+3)^3}\\
\\ &=\frac{(p+3)^3-(p-3)^3}{(p-3)^3(p+3)^3}\\
\\ &=\frac{(p+3)(p^2+6p+9)-(p-3)(p^2-6p-9)}{\left((p-3)(p+3)\right)^3}\\
\\ &=\frac{(p^3+6p^2+9p+3p^2+18p+27)-(p^3-6p^2-9p-3p^2+18p+27)}{\left(p^2-9\right)^3}\\
\\ &=\frac{18p^2+18p}{\left(p^2-9\right)^3}\\
\\ &=\frac{18p(p+1)}{\left(p^2-9\right)^3}\\
\end{align*}\)

Answer 9

now ?

Answer 10

now what can we do

Answer 11

thst the final answer, what u got

Answer 12

oh

Answer 13

though i may suggest another way to
\((p+3)^3-(p-3)^3\)
using a^3-b^3 formula would simplify it easier.....
rather than (a+b)^3 twice..but not much difference

Answer 14

what formula do you mean
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]?

Answer 15

yup, that one...

Answer 16

apply that to the denominator ,?

Answer 17

a-b=6

Answer 18

denominator is fine..

Answer 19

you entire answer is fine

Answer 20

i just suggested another way to simplify the numerator from \(
(p+3)^3-(p-3)^3\)

Answer 21

i not sure where you mean

Answer 22

\((p+3)^3-(p-3)^3 \\ (p+3-p+3)(p^2+6p+9+p^2-9+p^2-6p+9) \\ (6)(3p^2+9)\)
did i make any error ?

Answer 23

oh you mean like that

Answer 24

@UnkleRhaukus how di u get\( 18p^2+18p\)
i am getting
18p^2+54

Answer 25

your probably right

Answer 26

ohh..

Answer 27

\((p+3)^3 = (p+3)(p^2-3p+9)\)

Answer 28

\((p-3)^3=(p-3)(p^2+3p+9)\)

Answer 29

got it ?

Answer 30

thanks !

Answer 31

rest of the solution is correct
welcome ^_^