Find the general solution of the given differential equation: y''+4y'+6.25y=0
solution: The auxiliary equation is \begin{equation*}r^2+4r+6.25=0\end{equation*}
By the quadratic formula, the roots are \begin{equation*}r=\frac{-4\pm\sqrt{4^2 - 4(6.25)}}{2}=-2\pm\frac{3}{2}i.\end{equation*}
If the roots of the auxiliary equation are the complex numbers \begin{equation*}r=\alpha\pm i\beta\end{equation*} then the general solution of ay''+by'+cy=0 is\begin{equation*} y=e^{\alpha t} (C_1 cos\beta t +C_2sin\beta t)\end{equation*}
since\begin{equation*}\alpha=-2,\beta=\frac{3}{2}\end{equation*}
the general solution of the differential equation is \begin{equation*} y=e^{-2 t} (C_1 cos\frac{3}{2} t +C_2sin\frac{3}{2} t)\end{equation*}
