When $b^{2} - 4ac < 0$, the characteristic equation has complex roots $r = \alpha \pm \beta i$. The exponential $e^{rt}$ still solves the equation, but it is complex-valued. Polar form and Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ turn it into real solutions.
Key idea
If the roots are $\alpha \pm \beta i$ ($\beta > 0$), the general solution is $$y = e^{\alpha t}(c_1\cos\beta t + c_2\sin\beta t).$$ The real part $\alpha$ sets growth or decay; the imaginary part $\beta$ sets the angular frequency.
Why it works
By Euler's formula, $e^{(\alpha + \beta i)t} = e^{\alpha t}e^{i\beta t}$ $= e^{\alpha t}\cos\beta t + i\,e^{\alpha t}\sin\beta t$. Call this $u + iv$. The coefficients $a, b, c$ are real, so $L[u + iv] = L[u] + iL[v]$, and this is 0 only if both $L[u] = 0$ and $L[v] = 0$. So the real and imaginary parts are real solutions, with $W = \beta e^{2\alpha t} \ne 0$. (The other root $\alpha - \beta i$ gives the same pair again.)
Worked example: Decaying oscillation
Solve $y'' + 2y' + 5y = 0$, $y(0) = 1$, $y'(0) = 3$.
- $r = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$: $\alpha = -1$, $\beta = 2$.
- $y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$, so $y(0) = c_1 = 1$.
- $y'(0) = -c_1 + 2c_2 = 3$ (product rule at $t = 0$), so $c_2 = 2$.
Answer: $y = e^{-t}(\cos 2t + 2\sin 2t)$
Worked example: Pure imaginary roots
Solve $y'' + 9y = 0$, $y(0) = 2$, $y'(0) = -6$, and find the amplitude.
- $r^{2} + 9 = 0$: $r = \pm 3i$, so $\alpha = 0$, $\beta = 3$.
- $y = c_1\cos 3t + c_2\sin 3t$: $c_1 = 2$ and $3c_2 = -6$, so $c_2 = -2$.
- $c_1\cos 3t + c_2\sin 3t$ $= R\cos(3t - \delta)$ with $R = \sqrt{c_1^{2} + c_2^{2}} = 2\sqrt{2}$.
Answer: $y = 2\cos 3t - 2\sin 3t$, amplitude $2\sqrt{2}$
Worked example: Fractional real part
Find the general solution of $4y'' + 4y' + 17y = 0$.
- $r = \frac{-4 \pm \sqrt{16 - 272}}{8}$ $= \frac{-4 \pm 16i}{8} = -\frac{1}{2} \pm 2i$.
Answer: $y = e^{-t/2}(c_1\cos 2t + c_2\sin 2t)$
Common mistake
Writing $\cos((\alpha + \beta i)t)$ or $e^{\beta t}\cos\alpha t$. Only $\beta$, the imaginary part, goes inside the cosine and sine; the real part $\alpha$ goes in the exponential. A check: the solution must be real, and its envelope $e^{\alpha t}$ must shrink when $b > 0$ (positive damping).