# Laplacing ambigious equation, wants explanation:)

• Oct 17th 2009, 11:38 AM
kolowee9
Laplacing ambigious equation, wants explanation:)
Hello there, im trying out this from a workbook,I have no one to explain to me.

Could someone tell me the proper procedure of of laplacing this? it seems it is using the function of multiplication t, which is -F'(s) , but not as near to the answer(attached), could someone explain to me? especially the extra +(5/(s-1)) value which i dont think it comes from partial fraction. Thank you, kamsahamnida.(Bow)
• Oct 17th 2009, 05:15 PM
mr fantastic
Quote:

Originally Posted by kolowee9
Hello there, im trying out this from a workbook,I have no one to explain to me.

Could someone tell me the proper procedure of of laplacing this? it seems it is using the function of multiplication t, which is -F'(s) , but not as near to the answer(attached), could someone explain to me? especially the extra +(5/(s-1)) value which i dont think it comes from partial fraction. Thank you, kamsahamnida.(Bow)

There are several ways of doing this, including the following:

$LT\left[ t e^{t - 5} u(t - 5)\right] = \int_5^\infty t e^{t - 5} e^{-st} \, dt$

Make the substitution $u = t - 5$:

$= \int_0^\infty (u + 5) e^{u} e^{-s(u + 5)} \, dt = e^{-5s}\int_0^\infty (u + 5) e^{u} e^{-su} \, dt$

$= e^{-5s} \left( \int_0^\infty ue^{u} e^{-su} \, dt + 5 \int_0^\infty e^{u} e^{-su} \, dt \right)$

and the rest should be clear.

Spoiler:
In fact, since this is essentially just $e^{-5s} \left( LT[x e^x] + 5 LT[e^x] \right)$ you can just use tables and a Shifting Theorem to finish it off.