Having trouble taking the laplace of t^2 times the unit step function, any help is appreciated.

Question

mendicant_bias · Answer

I just don't know what steps I should take algebraically to get t^2 into the form of t-1; what i'm trying to do is take$$\mathcal{L}\left\{t^{2}\mathcal{U}(t-1)\right\}$$

mendicant_bias · Answer

Somehow, I need to change$$f(t)=t^2$$into the form$$f(t)=(t-1)^2$$

anonymous · Answer

ok

mendicant_bias · Answer

I'm thinking it might have something to do with$$(t-1)^{2}=t^2-2t+1$$

anonymous · Answer

yes

anonymous · Answer

that is right

anonymous · Answer

ok.....45328

mendicant_bias · Answer

I'd really rather you not waste my time. @satellite73

anonymous · Answer

so that wod be 54 6578

mendicant_bias · Answer

Alright. Now that that's out of the way.

mendicant_bias · Answer

@Concentrationalizing , Any idea?

mendicant_bias · Answer

Wait a second, I think I got it.

mendicant_bias · Answer

$$(t-1)^2=t^2-2t+1; \ \ \ (t-1)^2+2t-1=t^2.$$Now, the other term involving t that has been generated needs to be dealt with(?)

mendicant_bias · Answer

Subtract an additional one from the laplace and take the laplace of it as well?

anonymous · Answer

yes

mendicant_bias · Answer

$$\mathcal{L}\left\{[(t-1)^2+2(t-1)]\mathcal{U}(t-1)\right\}+\mathcal{L}\left\{1\right\}$$@Concentrationalizing , does this look right?

anonymous · Answer

yes

mendicant_bias · Answer

@Hero (or any mod), take a look at this, at least so you have an idea what might be a problem in the future.

anonymous · Answer

do you what me to fan you and medal you?

anonymous · Answer

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