# understanding another unit step function

So, my equation is $y'' + y = g(t)$

where $$g(t) = \begin{cases}t/2, & 0 \leq t < 6\\3, & t \geq 6\end{cases}$$

I understand completely the piece-wise function, and can graph it, no problems there. My problem arises when I need convert this to unit step notation. I've dealt with unit step functions that have an initial value of $0$, going to a certain value of $t$ but never with an initial value of $t/2$ or some other variable.

So I started off with that.. what if the initial value was 0? The step function from the above example would be $3u_{6}(t)$ but where does the $t/2$ come into play? Well, I need to add it in.. I think .. so we'd have $t/2 + 3u_{6}(t)$, but I also need to "remove it from the second function".. since I don't want $3 + t/2$ when $t\geq6$, so I would have..

$$g(t) =\frac{t}{2} + 3u_{6}(t) - \frac{3t}{2}u_{6}(t)$$

My question is, where do I start with this type of problem, am I thinking correctly? Is my solution correct?

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Almost. When $t < 6$, so $u_6(t) = 0$, your function is $t/2$ (good). When $t \ge 6$, so $u_6(t)=1$, your function is $3 - t$ (bad). You subtracted too much $t$.