Somebody told me that if i have something like this:
$x''(t) + x'(t) = -2x(t) + u$
$x(0) = 7$
and use laplace transform on it i will get
$s^2X(s) + sX(s) = -2X(s) + U(s)$
next i'm getting Transfer function form this above
$G(s) = \frac{1}{s^2+s-2}$
The main question is how it its possible when $x(0) = 7$?? The laplace formula on $x'$ is:
$laplace(x'(t)) = sX(s) - x(0)$
Why there is no $-x(0)$??
Edit: ----------------Second case -------------------------
$x'(t) = -2x(t) + 3sin(5t + \pi/6)$
$x(t) = ae^{-2t} + Asin(5t + \varphi)$
$x(0) = 7$
$t \ge 0$
The main goal is to get from this equation $A$ and $\varphi$
If i will be doing this like above i must take into account $x(0) = 7$ or this will something change? Is there any simple method to get $A$ and $\varphi$???
$$x''(t)+x'(t)=-2x(t)+u\Longleftrightarrow$$ $$\mathcal{L}_t\left[x''(t)+x'(t)\right]_{(s)}=\mathcal{L}_t\left[-2x(t)+u\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_t\left[x''(t)\right]_{(s)}+\mathcal{L}_t\left[x'(t)\right]_{(s)}=-2\mathcal{L}_t\left[x(t)\right]_{(s)}+\mathcal{L}_t\left[u(t)\right]_{(s)}\Longleftrightarrow$$ $$s^2x(s)-sx(0)-x'(0)+sx(s)-x(0)=-2x(s)+u(s)\Longleftrightarrow$$
So, we know that $x(0)=7$:
$$s^2x(s)-s\cdot7-x'(0)+sx(s)-7=-2x(s)+u(s)\Longleftrightarrow$$ $$s^2x(s)-7s-x'(0)+sx(s)-7+2x(s)=u(s)\Longleftrightarrow$$ $$x(s)\left[s^2+s+2\right]-7s-x'(0)-7=u(s)$$