# 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