Question

Find F(s) or f(t)
L{(3t+1)U(t-1)}

My answer: ((e^-s)/(s^2))+((4e^-s)/s)

Should I have a 3 in the front...((3e^-s)/(s^2))+((4e^-s)/s)?

Answer 1

Fyi, this involves translation on the t-axis (differential equations)

Answer 2

you need the 3

Answer 3

Zarkon, if you don't mind, can you show me how you worked out the problem?

Answer 4

\[L([3t+1]u(t-1))=\int\limits_{0}^{\infty}[3t+1]u(t-1)e^{-st}dt=\int\limits_{1}^{\infty}[3t+1]e^{-st}dt\]
=\[\int\limits_{1}^{\infty}3te^{-st}dt+\int\limits_{1}^{\infty}e^{-st}dt\]
=\[3\int\limits_{1}^{\infty}te^{-st}dt+\int\limits_{1}^{\infty}e^{-st}dt\]
now just do the integration (1st one by parts and the 2nd by substitution)
Note: it should read e^{-st}. the negative sign is not showing up ...at least not on my end

Answer 5

Ok, that's helpful, but I am supposed to use the Translation Theorem from Differential Equations: L{f(t-a)U(t-a)}=\[e ^{-as}F(s)\]

Answer 6

L(f(t-1)u(t-1))=e^{-as}F(s)

\[f(t)=3t+1=3(t-1)+4\]

so we need \[L(3t+4)=\frac{3}{s^2}+\frac{4}{s}\]

then multiply this by e^{-s} gives the answer

Answer 7

Not exactly the steps you use in D.E., but thanks for the help!