Calculate the laplace transform oft2u(t−2)I don't know how to manipulate&nbsp;t2&nbsp;in order for it to meet the form of the product between a function and a heaviside function.

The Heaviside function effective changes the lower limit of integration so the LT is∫2∞dtt2e−st=d2ds2∫2∞dte−st=d2ds2e−2ssTaking the derivative, the LT takes the form−dds[(2s+1s2)e−2s]=(4s+4s2+2s3)e−2s

Use {uc}(t)g(t)=e−csL{g(t+c)} from your table instead of {uc}(t)f({t−c})=e−csF(s). Then c=2 and g(t)=t2⇒g(t+c)=g(t+2)=(t+2)2=t2+4t+4soL{g(t+c)}=L{t2+4t+4}=2s3+4s2+4sThus, from {uc}(t)g(t)=e−csL{g(t+c)} in your table,L{t2u(t−2)}=e−2sL{g(t+2)}=e−2s(2s3+4s2+4s)

The Heaviside function effective changes the lower limit of integration so the LT is∫2∞dtt2e−st=d2ds2∫2∞dte−st=d2ds2e−2ssTaking the derivative, the LT takes the form−dds[(2s+1s2)e−2s]=(4s+4s2+2s3)e−2s

Use {uc}(t)g(t)=e−csL{g(t+c)} from your table instead of {uc}(t)f({t−c})=e−csF(s). Then c=2 and g(t)=t2⇒g(t+c)=g(t+2)=(t+2)2=t2+4t+4soL{g(t+c)}=L{t2+4t+4}=2s3+4s2+4sThus, from {uc}(t)g(t)=e−csL{g(t+c)} in your table,L{t2u(t−2)}=e−2sL{g(t+2)}=e−2s(2s3+4s2+4s)

"Calculate the laplace transform oft2u(t−2)I don't know how..." was answered by our community

Siliberti7cd

Answered question

2022-03-17

Calculate the laplace transform of
$t^{2} u (t - 2)$
I don't know how to manipulate $t^{2}$ in order for it to meet the form of the product between a function and a heaviside function.

Answer & Explanation

bridgedcuu

Beginner2022-03-18Added 7 answers

The Heaviside function effective changes the lower limit of integration so the LT is

\int_{2}^{\infty} d t t^{2} e^{- s t} = \frac{d^{2}}{d s^{2}} \int_{2}^{\infty} d t e^{- s t} = \frac{d^{2}}{d s^{2}} \frac{e^{- 2 s}}{s}

Taking the derivative, the LT takes the form

- \frac{d}{d s} [(\frac{2}{s} + \frac{1}{s^{2}}) e^{- 2 s}] = (\frac{4}{s} + \frac{4}{s^{2}} + \frac{2}{s^{3}}) e^{- 2 s}

stadfeste8ru

Beginner2022-03-19Added 9 answers

Use

{u_{c}} (t) g (t) = e^{- c s} L {g (t + c)}

from your table instead of

{u_{c}} (t) f ({t - c}) = e^{- c s} F (s)

. Then c=2 and

g (t) = t^{2} \Rightarrow g (t + c) = g (t + 2) = {(t + 2)}^{2} = t^{2} + 4 t + 4

L {g (t + c)} = L {t^{2} + 4 t + 4} = \frac{2}{s^{3}} + \frac{4}{s^{2}} + \frac{4}{s}

Thus, from

{u_{c}} (t) g (t) = e^{- c s} L {g (t + c)}

in your table,

L {t^{2} u (t - 2)} = e^{- 2 s} L {g (t + 2)} = e^{- 2 s} (\frac{2}{s^{3}} + \frac{4}{s^{2}} + \frac{4}{s})

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