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UnkleRhaukus

  • 3 years ago

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

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

  2. UnkleRhaukus
    • 3 years ago
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    or should i leave my solution as \[G(p)=\frac{1}{(p-3)^3}-\frac{1}{(p+3)^3}\]

  3. UnkleRhaukus
    • 3 years ago
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    ?

  4. Miyuru
    • 3 years ago
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    Don't know.. :)

  5. ChmE
    • 3 years ago
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    Your work is cut off pretty badly

  6. hartnn
    • 3 years ago
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    don't leave that answer the difference form, what simplification u have done was required, i think...

  7. Miyuru
    • 3 years ago
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    I can't see the rest of the steps @UnkleRhaukus how can i see that. ?

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

  9. hartnn
    • 3 years ago
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    now ?

  10. UnkleRhaukus
    • 3 years ago
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    now what can we do

  11. hartnn
    • 3 years ago
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    thst the final answer, what u got

  12. UnkleRhaukus
    • 3 years ago
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    oh

  13. hartnn
    • 3 years ago
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    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
    • 3 years ago
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    what formula do you mean \[a^3-b^3=(a-b)(a^2+ab+b^2)\]?

  15. hartnn
    • 3 years ago
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    yup, that one...

  16. UnkleRhaukus
    • 3 years ago
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    apply that to the denominator ,?

  17. hartnn
    • 3 years ago
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    a-b=6

  18. hartnn
    • 3 years ago
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    denominator is fine..

  19. hartnn
    • 3 years ago
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    you entire answer is fine

  20. hartnn
    • 3 years ago
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    i just suggested another way to simplify the numerator from \( (p+3)^3-(p-3)^3\)

  21. UnkleRhaukus
    • 3 years ago
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    i not sure where you mean

  22. hartnn
    • 3 years ago
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    \((p+3)^3-(p-3)^3 \\ (p+3-p+3)(p^2+6p+9+p^2-9+p^2-6p+9) \\ (6)(3p^2+9)\) did i make any error ?

  23. UnkleRhaukus
    • 3 years ago
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    oh you mean like that

  24. hartnn
    • 3 years ago
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    @UnkleRhaukus how di u get\( 18p^2+18p\) i am getting 18p^2+54

  25. UnkleRhaukus
    • 3 years ago
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    your probably right

  26. hartnn
    • 3 years ago
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    ohh..

  27. hartnn
    • 3 years ago
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    \((p+3)^3 = (p+3)(p^2-3p+9)\)

  28. hartnn
    • 3 years ago
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    \((p-3)^3=(p-3)(p^2+3p+9)\)

  29. hartnn
    • 3 years ago
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    got it ?

  30. UnkleRhaukus
    • 3 years ago
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    thanks !

  31. hartnn
    • 3 years ago
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    rest of the solution is correct welcome ^_^

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