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I was given looking at one of the examples in my textbook and it took this laplace transform $L\{u(t-3)(t^2)\}$ and turned it into this $L\{u(t-3)[(t-3)^2+6(t-3)+9]\}$ in the next step.

I'm wondering how did $t^2$ become $(t-3)^2+6(t-3)+9$ I get you can replace $t^2$ with $(t-3+3)^2$ to use the $2^{nd}$ shifting theorem later but i don't know how to get the 6(t-3)+9 that came after the $(t-3)^2$.

iadvd
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a wee little worker
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1 Answers1

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Note that $(x+y)^2 = x^2+2xy+y^2$ for any numbers $x$ and $y$.

Using that identity for $x = t-3$ and $y = 3$, you get $t^2 = ((t-3)+3)^2 = (t-3)^2+2 \cdot (t-3) \cdot 3 + 3^2 = (t-3)^2+6(t-3)+9$.

JimmyK4542
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