Concept:
Case #3 deals with the situation when our characteristic equation has complex roots. In order to deal with such a case, we will make use of Euhler's Identity
Euhler's Identity:
$$e^{i\theta} = cos\theta + isin\theta$$
(Ex 1) Solve the following initial value problem for the given 2nd order homogeneous differential equation:
$$y''+4y'+5y=0 $$
...where:
$$y(0)=1 \;,\;y'(0)=0 $$
Start with making our substitutions:
$$y=e^{rx} $$
$$y'=re^{rx} $$
$$y''=r^2e^{rx} $$
...and with these substitutions we obtain a characteristic equation of:
$$r^2e^{rx}+4re^{rx}+5e^{rx}=0 $$
$$r^2+4r+5=0 $$
In order to find our roots, we will use the quadratic formula:
$$r=\frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
$$r=\frac{-4 \pm \sqrt{16-4(1)(5)}}{2} $$
$$r=\frac{-4 \pm \sqrt{-4}}{2} $$
$$r=\frac{-4 \pm \sqrt{4(-1)}}{2} $$
$$r=\frac{-4 \pm 2\sqrt{-1}}{2} $$
$$r=\frac{-4 \pm 2i}{2} $$
$$r=-2 \pm i $$
Taking only the positive sign of our above expression and back-substituting for y we obtain:
$$y=e^{(-2+i)x} $$
$$y=e^{-2x+ix} $$
$$y=e^{-2x}e^{ix} $$
Now we use Euhler's identity to rewrite the above expression as:
$$y=e^{-2x}[cos(x)+isin(x)] $$
We then disregard the imaginary "i" and obtain:
$$y=e^{-2x}cos(x)+e^{-2x}sin(x) $$
...which is the fundamental set:
$$e^{-2x}cos(x) \;,\; e^{-2x}sin(x) $$
Therefore, the general solution to our given differential equation is represented by:
$$y(x) = C_1e^{-2x}cos(x) + C_2e^{-2x}sin(x) \qquad(1)$$
Using our given initial conditions, we proceed to solve the initial value problem:
$$y(0) = C_1e^{-2(0)}cos(0) + C_2e^{-2(0)}sin(0)=1 $$
$$C_1 =1 $$
$$y'(x) = C_1[-2e^{2x}cos(x)-e^{-2x}sin(x)] $$
$$\qquad \quad + C_2[-2e^{-2x}sin(x)+e^{-2x}cos(x)] = 0 $$
$$C_1[-2(1)(1)-1(0)] + C_2[-2(0)+1(1)] = 0 $$
$$-2C_1+C_2=0 \;,\; where\; C_1 = 1 $$
$$-2 + C_2 = 0 $$
$$C_2 = 2 $$
Using our recently obtained values for C1 and C2, and plugging them into our general solution, we obtain the particular solution (yp) to our initial value problem:
$$y_p = e^{-2x}cos(x) + 2e^{-2x}sin(x) $$