Solving Second Order Differential Equations with Complex Roots

fact
Remember that the "characteristic equation" for a linear second order differential equation $$ay'' + by' + cy = 0$$ is given by: $$ar^2 + br + c = 0$$
fact
If the roots to the characteristic equation are complex $$\lambda \pm \mu i$$ then the two solutions to the differential equation are: $$y_1 = e^{(\lambda + \mu i)t}$$ and $$y_2 = e^{(\lambda - \mu i)t}$$
Using Euler's formula we can rewrite these complex exponentials in terms of sines and cosines
fact
The general solution to a second order differential equation with complex roots ($$\lambda \pm \mu i$$) is given by: $$c_1e^{\lambda t}\cos(\mu t) + c_2e^{\lambda t}\sin(\mu t)$$
derivation
Euler's formula tells us that $$e^{ix} = \cos(x) + i\sin(x)$$ $$y_1 = e^{(\lambda + \mu i)t} = e^{\lambda t}e^{\mu i t} = e^{\lambda t}\left( \cos(\mu t) + i\sin(\mu t) \right)$$ $$y_2 = e^{(\lambda - \mu i)t} = e^{\lambda t}e^{-\mu i t} = e^{\lambda t} \left( \cos(\mu t) - i\sin(\mu t) \right)$$ $$= e^{\lambda t}\left( \cos(\mu t) + i\sin(\mu t) + \cos(-\mu t) + i\sin(-\mu t) \right)$$ $$= e^{\lambda t}2 \cos(\mu t)$$ $$y_1 - y_2 = e^{\lambda t}2i\sin(\mu t)$$ Now with the correct choice of constants $$c_1$$ and $$c_2$$ we can get rid of those pesky coefficients and be left with the two base solutions: $$u(t) = e^{\lambda t}\cos(\mu t)$$ and $$v(t) = e^{\lambda t}\sin(\mu t)$$ Which when combined give us the general solution: $$y(t) = c_1 e^{\lambda t}\cos(\mu t) + c_2e^{\lambda t}\sin(\mu t)$$ We don't need to worry about the $$2$$ before the cosine or the $$2i$$ before the sine as we just assume the constants $$c_1$$ and $$c_2$$ are picked to remove them anyway.
practice problems