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

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

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Other answers:

Don't know.. :)
Your work is cut off pretty badly
don't leave that answer the difference form, what simplification u have done was required, i think...
I can't see the rest of the steps @UnkleRhaukus how can i see that. ?
\(\begin{align*} \\\text{(b)} &g(t)=t^2\sinh(3t)\\ \\ &G(p)=\mathcal L\{t^2\sinh(3t)\}\\ \\ &=\int\limits_0^\infty t^2\sinh(3t)e^{-pt}\text dt\\ \\ &=\int\limits_0^\infty t^2\left(\frac{e^{3t}-e^{-3t}}{2}\right)e^{-pt}\text dt\\ \\ &=\frac12\int\limits_0^\infty t^2\left(e^{3t}-e^{-3t}\right)e^{-pt}\text dt\\ \\ &=\frac12\left[\int\limits_0^\infty t^2e^{-(p-3)}\text dx-\int\limits_0^\infty t^2e^{-(p+3)}\text dt\right]\\ \\ &=\frac12\left[\frac{2!}{(p-3)^3}-\frac{2!}{(p+3)^3}\right]\\ \\ &=\frac{1}{(p-3)^3}-\frac{1}{(p+3)^3}\\ \\ &=\frac{(p+3)^3-(p-3)^3}{(p-3)^3(p+3)^3}\\ \\ &=\frac{(p+3)(p^2+6p+9)-(p-3)(p^2-6p-9)}{\left((p-3)(p+3)\right)^3}\\ \\ &=\frac{(p^3+6p^2+9p+3p^2+18p+27)-(p^3-6p^2-9p-3p^2+18p+27)}{\left(p^2-9\right)^3}\\ \\ &=\frac{18p^2+18p}{\left(p^2-9\right)^3}\\ \\ &=\frac{18p(p+1)}{\left(p^2-9\right)^3}\\ \end{align*}\)
now ?
now what can we do
thst the final answer, what u got
oh
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
what formula do you mean \[a^3-b^3=(a-b)(a^2+ab+b^2)\]?
yup, that one...
apply that to the denominator ,?
a-b=6
denominator is fine..
you entire answer is fine
i just suggested another way to simplify the numerator from \( (p+3)^3-(p-3)^3\)
i not sure where you mean
\((p+3)^3-(p-3)^3 \\ (p+3-p+3)(p^2+6p+9+p^2-9+p^2-6p+9) \\ (6)(3p^2+9)\) did i make any error ?
oh you mean like that
@UnkleRhaukus how di u get\( 18p^2+18p\) i am getting 18p^2+54
your probably right
ohh..
\((p+3)^3 = (p+3)(p^2-3p+9)\)
\((p-3)^3=(p-3)(p^2+3p+9)\)
got it ?
thanks !
rest of the solution is correct welcome ^_^

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