Consider the ODE
y''(t) + T\cdot y'(t) +D\cdot y(t) = 0.
Determine the general solution y with y(t) and constants
A und B, where A should precede B in the sense that A.
y(t) =
e^{-T/2*t} (A * \sqrt{-K/4}*t)
+ B * \sin(\sqrt{-K/4} *t))
Use the characteristic equation to get the general solution of the ODE
\lambda^2 + T\cdot \lambda +D = 0.
The two (complex) roots of the quadratic equation
are \lambda_{1,2} = \dfrac12 (-T \pm formattedSquareRootOf(-K) i).
Thus, the general solution is
y(t) = e^{(-T/2) t} \left(A \cos(\beta t)
+ B \sin(\beta t) \right) with
\beta = \dfrac{formattedSquareRootOf(-K)}2.