## UnkleRhaukus 3 years ago Find the Laplace transform of the following function $g(t)=t^2\sinh(3t)$

1. UnkleRhaukus

\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*}

2. UnkleRhaukus

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

3. UnkleRhaukus

?

4. Miyuru

Don't know.. :)

5. ChmE

6. hartnn

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

7. Miyuru

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

8. hartnn

\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*}

9. hartnn

now ?

10. UnkleRhaukus

now what can we do

11. hartnn

thst the final answer, what u got

12. UnkleRhaukus

oh

13. hartnn

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

14. UnkleRhaukus

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

15. hartnn

yup, that one...

16. UnkleRhaukus

apply that to the denominator ,?

17. hartnn

a-b=6

18. hartnn

denominator is fine..

19. hartnn

20. hartnn

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

21. UnkleRhaukus

i not sure where you mean

22. hartnn

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

23. UnkleRhaukus

oh you mean like that

24. hartnn

@UnkleRhaukus how di u get$$18p^2+18p$$ i am getting 18p^2+54

25. UnkleRhaukus

26. hartnn

ohh..

27. hartnn

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

28. hartnn

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

29. hartnn

got it ?

30. UnkleRhaukus

thanks !

31. hartnn

rest of the solution is correct welcome ^_^