# Laplace transformation with $(t+3)^2$

The problem says that: $$F(s) = \frac{-2 \times {(s)}^{-1/2}}{1}$$

I need to find the laplace transformation of $${e}^{-8t} \times f(t) \times ({t+3})^{2}$$

I know that if it was $$({t})^{2}$$ instead of $$({t+3})^{2}$$, I would have to: 1st) substitute s for s+8 (in my country we say translation, right?), than I would have to find the second derivate of it. I've already done the 1st step (s+8 thing) and got $$F(s) = \frac{-2 \times {(s+8)}^{-1/2}}{1}$$ What am I supposed to do now? Sorry for my english, I hope you understand my doubt.

• What is $f(t)$? Mar 23, 2021 at 1:17
• You can use linearity. $(t+3)^2 = t^2 + 6t + 9$
– Joe
Mar 23, 2021 at 1:27
• f(t) is the function which laplace transformation is $$F(s) = \frac{-2 \times {(s)}^{-1/2}}{1}$$ Mar 23, 2021 at 1:27

To solve this question, you need to use a combination of some properties of the Laplace Transform: linearity, frequency shift and frequency differentiation. Let's remember them first.


Let $$g(t) = e^{-8t} \cdot f(t) \cdot(t+3)^2$$. Rewriting the expression, in order to use the above properties, we have:

$$g(t) =\underbrace{t^2e^{-8t}f(t)}_{\text{(I)}} \,\,+ \,\,\underbrace{6te^{-8t}f(t)}_{\text{(II)}} \,\,+ \,\, \underbrace{9e^{-8t}f(t)}_{\text{(III)}}$$

For $$\text{(I)}$$, we use frequency shift combined with frequency differentiation. For $$\text{(II)}$$, we use the previous ones plus linearity. And for $$\text{(III)}$$ we use frequency shift and linearity. Then $$G(s)$$ is:

$$G(s) = (-1)^2 F^{(2)}(s+8) \,\, + \,\, 6\cdot (-1)^1 F^{(1)}(s+8) \,\, + \,\, 9F(s+8)$$

Since $$F(s) = -2\cdot (s)^{-1/2}$$, we have:

\begin{alignat}{1} G(s) &= -\frac{-3/2}{\sqrt{s+8}\cdot (s+8)^2}-\frac{6}{\sqrt{s+8}\cdot (s+8)} -\frac{18}{\sqrt{s+8}} \\[6pt] &= -\frac{-3/2 \cdot (12s^2+196s+801)}{\sqrt{s+8}\cdot (s+8)^2} \end{alignat}

• Thank you, @Vinicius ACP!!! Mar 25, 2021 at 16:01