Trying to find a general solution to a simplified equation of inertial movement in a non-inertial reference frame. For simplicity I'm restricting motion to the plane of rotation and making sure the vector of rotation is constant on the Z-axis. My issue is that I only get one solution, which should be the motion of a stationary object, and I can't seem to find the term to impart an initial velocity. What am I missing?
From the physics equation for the apparent acceleration of an inertial object in a rotating reference frame, I get: $$\frac{d^2 \vec r}{d t^2} = -2 \vec \omega \times \frac{d \vec r}{d t} - \vec \omega \times (\vec \omega \times \vec r)$$
Simplifying this, and looking at the components in the x-y plane. The differential equations I end up with are: $$ \frac{d^2 x}{d t^2} = 2 \omega \frac{d y}{d t} + \omega^2 x $$ $$\frac {d^2 y}{d t^2} = - 2 \omega \frac{d x}{d t} + \omega^2 y $$
I start solving this by introducing S = x + i y: $$\frac{d^2 S}{d t^2} = \frac {d^2 x}{d t^2}+ i \frac {d^2 y}{d t^2}$$ thus $$\frac{d^2 S}{dt^2} + 2 i \omega \frac{d S}{d t} - \omega^2 S = 0$$ By defining S as an exponential function, I get this solution: $$S = S_0 e^{-i \alpha t}$$ $$-\alpha^2 + 2 \omega \alpha - \omega^2=0: \alpha = \omega$$
This works just fine as one solution, but as a second-order differential equation, shouldn't there be a second equation? And physically this looks fine for the apparent position of a stationary object according to the rotating reference frame, but how do I get the term corresponding to velocity, for an object that has a constant velocity in the inertial reference frame?