- 41

- 0

[tex] L[/tex] is Laplace Transform operator.

Question is:

Let [tex] f(t) = t^2[/tex]. Derive [tex]L[f][/tex] from [tex]L[1][/tex]

So I know [tex]f(1) =1[/tex] and [tex] L[1] = \frac{1}{s}[/tex]

Carrying out the Transform...

[tex]L[f] = \int_{0}^{\infty} e^{-st}t^2 dt[/tex]

Integration by parts [tex] u = t^2, dv = e^{-st} dt [/tex]

I do the integration, the uv terms go to zero. after I clean up I get

[tex] - \int_{0}^{\infty} \frac{e^{-st}}{-s} 2t\ dt[/tex]

Doing IBP again. [tex] u = t,\ dv = e^{-st} dt[/tex] Rinse and repeat.

[tex] - \frac{2}{s} \int_{0}^{\infty} \frac{e^{-st}}{-s} (1) dt \\

= - \frac{2}{s} \left( \frac{e^{-st}}{s^2}\right)_{0}^{\infty} = - \frac{2}{s^3} [/tex]

Ok. So is that right? I don't think the negative is right at the end. But what the bigger thing that worries me is how do I use L[1] in this? Or is that a different approach?

Question is:

Let [tex] f(t) = t^2[/tex]. Derive [tex]L[f][/tex] from [tex]L[1][/tex]

So I know [tex]f(1) =1[/tex] and [tex] L[1] = \frac{1}{s}[/tex]

Carrying out the Transform...

[tex]L[f] = \int_{0}^{\infty} e^{-st}t^2 dt[/tex]

Integration by parts [tex] u = t^2, dv = e^{-st} dt [/tex]

I do the integration, the uv terms go to zero. after I clean up I get

[tex] - \int_{0}^{\infty} \frac{e^{-st}}{-s} 2t\ dt[/tex]

Doing IBP again. [tex] u = t,\ dv = e^{-st} dt[/tex] Rinse and repeat.

[tex] - \frac{2}{s} \int_{0}^{\infty} \frac{e^{-st}}{-s} (1) dt \\

= - \frac{2}{s} \left( \frac{e^{-st}}{s^2}\right)_{0}^{\infty} = - \frac{2}{s^3} [/tex]

Ok. So is that right? I don't think the negative is right at the end. But what the bigger thing that worries me is how do I use L[1] in this? Or is that a different approach?

Last edited: