# Recitation 6

Given a 2nd order linear homogeneous differential equation ay'' + by' + cy = 0, the characteristic equation ar^2 + br + c = 0 may have two complex roots r_1, r_2 = \lambda \pm i\mu. In this case, we would still like to write the solution as y = c_1 e^{r_1 x} + c_2 e^{r_1 x}. But what does it mean to raise e to a complex number?

Taylor expansion of e^t = \sum_{n=0}^\infty \frac{t^n}{n!} gives the clue: e^{it} = \sum_{n=0}^\infty \frac{(it)^n}{n!} = \sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{(2n)!} + i\sum_{n=0}^\infty\frac{(-1)^nt^{2n+1}}{(2n+1)!} = \cos t + i \sin t.

Using Euler’s formula e^{it} = \cos t + i\sin t, we can rewrite y = c_1 e^{r_1 x} + c_2 e^{r_1 x}, where r_1, r_2 = \lambda \pm i\mu, as a real-valued solution y = e^{\lambda x}(C_1\cos\mu x + C_2 \sin \mu x).

Example 1 Solve the IVP 16y'' - 8y' + 145y = 0, y(0) = -2, y'(0) = 1.

Solution The characteristic equation 16r^2 - 8r + 145 has complex roots r = 1/4 \pm 3i. The general solution of the DE is y = e^{x/4}\left(c_1\cos 3x + c_2\sin 3x\right). Thus y' = \frac{1}{4}e^{x/4}\left(c_1\cos 3x + c_2\sin 3x\right) + e^{x/4}\left(-3c_1\sin 3x + 3c_2\cos 3x\right). The initial conditions imply c_1 = -2, c_1/4 + 3c_2 = 1 and so c_1 = -2, c_2 = 1/2.

Example 2 Solve the IVP 3y'' - y' + 2y = 0, y(0) = 2, y'(0) = 0.

Solution The characteristic equation 3r^2 - r + 2 = 0 has complex roots r = \frac{1}{6} \pm \frac{\sqrt{23}}{6}i. The general solution is y = e^{x/6}\left(c_1\cos\frac{\sqrt{23}}{6}x + c_2\sin\frac{\sqrt{23}}{6}x\right). The initial conditions imply c_1 = 2, \frac{1}{6}c_1 + \frac{\sqrt{23}}{6}c_2 = 0, and so c_1 = 2, c_2 = \frac{-2}{\sqrt{23}}.

Example 3 Solve the IVP 5y'' + 2y' + 7y = 0, y(0) = 2, y'(0) = 1.

Solution The characteristic equation 5r^2 + 2r + 7 = 0 has complex roots r = -\frac{1}{5} \pm \frac{\sqrt{34}{5}}i. The general solution is y = e^{-x/5}\left(c_1\cos\frac{\sqrt{34}}{5}x + c_2\sin\frac{\sqrt{34}}{5}x\right). The initial conditions give c_1 = 2, c_2 = \frac{7}{\sqrt{34}}.