Second Shifting Theorem

Question

I need to find the Laplace transform of the following function $\frac{1}{4} tu(t-7)$ using the second shifting theorem

My working is as follows

$\frac{1}{4} tu(t-7) = e^{-7s} L(\frac{1}{4}(t+7))$

$= e^{-7s} (\frac{1}{4s^2}+ \frac{7}{4s})$

Is this correct or do I need to do more working/completely find a new answer

Answer 1

$$f(t)=\frac{1}{4} tu(t-7)$$ Rewrite it as: $$f(t)=\frac{1}{4}( (t-7)u(t-7)+7u(t-7))$$ $$f(t)=\frac{1}{4} (t-7)u(t-7)+\frac{7}{4}u(t-7)$$ Apply the Shift Theorem. $$F(s)=\dfrac {e^{-7s}} {4s^2}+\dfrac {7e^{-7s}}{4s}$$ $$F(s)=\dfrac {e^{-7s}} {4s}\left(\dfrac 1s+ 7\right)$$