Solving Second Order Differential Equations with Complex Roots

Remember that the "characteristic equation" for a linear second order differential equation \(ay'' + by' + cy = 0\) is given by: $$ ar^2 + br + c = 0 $$
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
The general solution to a second order differential equation with complex roots (\(\lambda \pm \mu i\)) is given by: $$ c_1e^{\lambda t}\cos(\mu t) + c_2e^{\lambda t}\sin(\mu t) $$
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