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Hi I have been trying to do this Laplace Transform and cant seem to figure it out and was wondering if someone could point me in the right direction; here it is:

$$L_t{(u(t-2)(2t^2-6t+5)})$$

What I tried was:

$$L_t{(u(t-2)(2t^2-6t+5)})$$

$$=e^{-2s}L_t({2t^3-10t^2+17t-10})$$

$$=e^{-2s}(\cfrac{12}{s^4} - \cfrac{20}{s^3} + \cfrac{17}{s^2} - \cfrac{10}{s})$$

But this is wrong. Can anyone tell what I have done wrong and help me figure it out.?

Thanks.

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  • $\begingroup$ The "unit step function" is given by $$u(t) = \begin{cases} 0 &, t < 0\\ 1 &, t > 0\end{cases},$$ with the value for $t = 0$ being one of $0,1,\frac{1}{2}$? $\endgroup$ – Daniel Fischer Jul 3 '14 at 13:16
  • $\begingroup$ $ 2 t^2 - 6 t +5 = A (t-2)^2 +B(t-2) + C$, find $A$, $B$ and $C$ then you can use the time delay property. $\endgroup$ – oholmer Jul 3 '14 at 13:18
  • $\begingroup$ $u(t-2)$ is the unit step function, $(t-2)$ is the input, and is not a factor, you can't multiply it to that quadratic function. $\endgroup$ – Shuhao Cao Jul 3 '14 at 13:24
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By the formula $L(u(t-a)f(t))=e^{-as}Lf(t+a)$ we have
\begin{aligned} L(U(t-2)(2t^2-6t+5)) &=e^{-2s}L((2(t+2)^2-6(t+2)+5)) \\ &=e^{-2s}(\frac{4}{s^3}+\frac{2}{s^3}+\frac{1}{s}) \\ \end{aligned}

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