Second order constant coefficients (II)

Problem:

$ y'' + y' + y = 0 $

Solution:

The characteristic polynomial is $ x^2 + x + 1 = 0 $, the solution is $ \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{3} i}{2} $

Therefore the answer is

$ y = Ae^{\frac{-1 + \sqrt{3} i}{2}x} + Be^{\frac{-1 - \sqrt{3} i}{2}x} $

Separating out the real and imaginary parts, we get

$ y = Ae^{\frac{-1}{2}x} e^{\frac{\sqrt{3} i}{2}x} + Be^{\frac{-1}{2}x}e^{\frac{- \sqrt{3} i}{2}x} $

Next we can apply the Euler's formula

$ y = Ae^{\frac{-1}{2}x} (\cos\frac{\sqrt{3}}{2}x + i\sin\frac{\sqrt{3}}{2}x) + Be^{\frac{-1}{2}x}(\cos\frac{\sqrt{3}}{2}x - i\sin\frac{\sqrt{3}}{2}x) $

Grouping the real and imaginary gives:

$ y = Ae^{\frac{-1}{2}x} \cos\frac{\sqrt{3}}{2}x  + Be^{\frac{-1}{2}}\cos\frac{\sqrt{3}}{2}x + Ae^{\frac{-1}{2}}\sin(\frac{\sqrt{3}}{2}x)i - Be^{\frac{-1}{2}x}\sin(\frac{\sqrt{3}}{2}x)i $

Factorize, we get:

$ y = (A + B) e^{\frac{-1}{2}x} \cos\frac{\sqrt{3}}{2}x + (A-B)e^{\frac{-1}{2}}\sin(\frac{\sqrt{3}}{2}x)i $

There are two ways we can make the expression real, either we set $ A = B $, that get rid of the imaginary part, or we can set $ A = -B $ be a pure imaginary number, that essentially get rid of the real part and make the imaginary part real. That gives the answer as:

$ y = C e^{\frac{-1}{2}x} \cos\frac{\sqrt{3}}{2}x + D e^{\frac{-1}{2}}\sin\frac{\sqrt{3}}{2}x $