I'm solving past exam questions in preparation for an Applied Mathematics course. I came to the following exercise, which poses a single difficulty. If it's any indication of difficulty, the exercise is only Part 2-B of the sheet, graded for 10%
Solve the following initial value problem using Laplace transformation: $y''(t)+6y'(t)+5y(t)=u(t-5)$ with $y(0)=1, y'(0)=1$ The exercise also gives a table for Laplace transformations, and notably: ${\cal L}\{ u(t-a)f(t-a)\} = e^{-ts}F(s), a\geq 0$
I understand that $u(t-5)$ is the unit step function, but what is its ${\cal L}$ transform?
If I begin getting the transforms, I get ${\cal L}\{ y''\} + 6{\cal L}\{ y'\} +5{\cal L}\{ y \} = {\cal L}\{ u(t-5)\}$
The first part of the equation poses no problem, but I do not know how to transform the unit step function, and I can't use the given transformation in the exercise notes as I do not have a $f(t-a)$ next to it.