Question

 Find the general solution of the equation y^''− 4y^'+ 13y = 0.

Answer 1

$\text{Given } y'' - 4y'+13y = 0 \\ \text{The auxilliary equation is of the form } m^2 -4m + 13 = 0 \text{, using the quadratic formula we have } \\ m = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i \\ \text{Therefore the general solution of the given differential is } \\ y(x) = \mathrm{e}^{2x} (A \cos 3x + B \sin 3x) \text{ where A and B are constants.}$