Auxiliary Equation: Two Complex Roots
The third, final case for the auxiliary equation (again, see the tutorial before) is having two complex roots. For this to happen, the quadratic auxiliary equation \( r^2 + \alpha r + \beta = 0 \) will have a negative discriminant:
\begin{equation}
\Delta = \alpha^2 -4\beta < 0 \tag{1}
\end{equation}
and thus the complex roots are
\begin{align}
r_{\pm} &= \frac{-\alpha \pm \sqrt{\alpha^2 -4\beta}}{2} \\
&= \frac{-\alpha}{2} \pm \frac{\sqrt{4\beta-\alpha^2}}{2}i \tag{2}
\end{align}
Euler’s Formula
The above suggests that the two solutions to the ODE in question will be
\begin{align}
y_1 , y_2 = e^{(-\alpha/2 \pm (\sqrt{4\beta-\alpha^2}/2)i)x} \tag{3}
\end{align}
To handle the complex exponential, we will make use of the famous Euler’s Formula:
\begin{equation}
e^{i\theta} = \cos \theta + i \sin \theta \tag{4}
\end{equation}
So that (3) becomes
\begin{align}
y_1 , y_2 &= e^{-(\alpha/2) x}(\cos ((\sqrt{4\beta-\alpha^2}/2)x) \pm i \sin ((\sqrt{4\beta-\alpha^2}/2)x)) \\
&= e^{-(\alpha/2) x}(\cos (qx) \pm i \sin (qx)) \tag{5}
\end{align}
where we set \( q = \sqrt{4\beta-\alpha^2}/2 \).
The problem with this is that the two proposed solutions are complex-valued while the original ODE has real coefficients, and it will be more desirable to have real solutions instead. Now, it is instructive to denote the ODE using an operator such that \( \mathcal{L}[y] = 0 \), and notice that we can simply decompose (5) into real and complex parts:
\begin{align}
\mathcal{L}[e^{-(\alpha/2) x}(\cos (qx) \pm i \sin (qx))] &= 0 \\
\mathcal{L}[e^{-(\alpha/2) x}\cos (qx)] \pm i \mathcal{L}[e^{-(\alpha/2) x}\sin (qx)] &= 0 \tag{6}
\end{align}
Since \( \mathcal{L} \) is a real operator as the ODE has real coefficients, by equating the real and complex parts to zero, we must have \( \mathcal{L}[e^{-(\alpha/2) x}\cos (qx)] = \mathcal{L}[e^{-(\alpha/2) x}\sin (qx)] = 0 \), and hence we now have two real-valued solutions instead:
\begin{align}
\left\{\begin{aligned}
y_1 &= e^{-(\alpha/2) x}\cos (qx) \\
y_2 &= e^{-(\alpha/2) x}\sin (qx)
\end{aligned}\right. \tag{7}
\end{align}
If we denote the two complex roots for the auxiliary equation as \( r_{\pm} = p \pm iq \), then by the same construction the solutions will be \( y_1, y_2 = e^{px}\cos (qx), e^{px}\sin (qx) \). We can subsequently check their linear independence by the Wronskian test (see this tutorial):
\begin{align}
&\quad W[y_1, y_2](x) \\
&= \begin{vmatrix}
e^{px}\cos (qx) & e^{px}\sin (qx) \\
pe^{px}\cos (qx) -qe^{px}\sin (qx) & pe^{px}\sin (qx) + qe^{px}\cos (qx)
\end{vmatrix} \\
&= (pe^{2px}\sin (qx)\cos (qx) + qe^{2px}\cos^2 (qx)) \\
&\quad -(pe^{2px}\sin (qx)\cos (qx) -qe^{2px}\sin^2 (qx))\\
&= qe^{2px} \neq 0 \tag{8}
\end{align}
Example
Solve the ODE
\begin{equation}
\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0 \tag{9}
\end{equation}
with the initial conditions of \( y(0) = -1, y'(0) = 3 \).
The auxiliary equation has
\begin{align}
r^2 + 2r + 5 &= 0 \\
r &= \frac{-2 \pm \sqrt{(2)^2 – 4(1)(5)}}{2} \\
&= -1 \pm \frac{\sqrt{-16}}{2} = -1 \pm \frac{4i}{2} = -1 \pm 2i \tag{10}
\end{align}
as its two complex roots. By the remark after (7), the two real-valued solution bases are \( y_1 = e^{-x}\cos (2x), y_2 = e^{-x} \sin (2x) \), hence \( y’_1 = -2e^{-x}\sin(2x) -e^{-x}\cos(2x), y’_2 = 2e^{-x}\cos(2x) -e^{-x}\sin(2x) \). Substituting the I.C. into their linear combination \(c_1 y_1 + c_2 y_2 \), we have
\begin{align}
&\left\{
\begin{aligned}
c_1(1) + c_2(0) &= -1 \\
c_1(-1) + c_2(2) &= 3 \\
\end{aligned}\right. \\
& \Rightarrow c_1 = -1, c_2 = 1 \\
& \Rightarrow y = -e^{-x}\cos(2x) + e^{-x}\sin(2x) \tag{11}
\end{align}
Exercise
Solve the second-order ODE
\begin{equation}
\frac{d^2y}{dx^2} + 12y = 0 \tag{12}
\end{equation}
where the I.C. consists of \( y(0) = 1, y'(0) = 2 \).
Answer
The auxiliary equation is simply \( r^2 + 12 = 0 \) and hence the complex roots are easily seen to be \( r_\pm = 2\sqrt{3}i \). The two valid solutions are \( y_1 = \cos(2\sqrt{3}x) \) and \( y_2 = \sin(2\sqrt{3}x) \), so the general solution will take the form of \( y = c_1 \cos(2\sqrt{3}x) + c_2\sin(2\sqrt{3}x) \). Substituting the I.C. readily yields \( c_1 = 1, c_2 = 2/(2\sqrt{3}) = 1/\sqrt{3} \), i.e. \( y = \cos(2\sqrt{3}x) + \sin(2\sqrt{3}x)/\sqrt{3} \).
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