Complex and Repeated Roots

Complex characteristic roots describe every underdamped mechanical oscillator: a mass–spring–damper system $mx''+cx'+kx=0$ has complex roots exactly when damping is light, giving the familiar decaying-sine motion seen in shock absorbers and seismographs. In electrical engineering, the same mathematics governs RLC circuits, where $\alpha$ is the decay rate of transient current and $\beta$ is the ringing frequency; resonance and filter design in signal processing depend on locating these roots in the complex plane. Repeated real roots correspond to critical damping, the precise boundary between oscillatory and non-oscillatory return to equilibrium, used deliberately in the design of door closers, analog meter needles, and control systems tuned to settle without overshoot as fast as possible. In control theory, the location of characteristic (here, closed-loop pole) roots in the complex plane — real part negative for stability, imaginary part for oscillation frequency — is the entire content of root-locus and pole-placement design, a direct descendant of this classification. In quantum mechanics, the time-independent Schrödinger equation for bound states produces eigenvalue problems whose characteristic-type analysis (though for boundary value problems, not IVPs) still hinges on the same real/complex root dichotomy. The reduction-of-order technique used to build $te^{rt}$ generalizes directly to constructing missing solutions for Higher-Order Linear Equations with roots of higher multiplicity, and both root types reappear as the "homogeneous part" one must add to any particular solution found by the Method of Undetermined Coefficients.

Ask the tutor: How does this classification of characteristic roots extend to third- and fourth-order constant-coefficient equations with mixed real, repeated, and complex roots?

Start from what already works: when the characteristic equation has two distinct real roots $r_1, r_2$, the functions $e^{r_1 t}$ and $e^{r_2 t}$ are visibly independent — one grows or decays at one rate, the other at a different rate, and no constant multiple of one can equal the other. The whole difficulty of complex and repeated roots is that this visible independence disappears. When roots are complex, $e^{r_1t}$ and $e^{r_2t}$ are complex-valued functions of a real variable, and we need genuinely real solutions to describe a real physical system. When roots collide, $e^{r_1t}$ and $e^{r_2t}$ become the same function, and we have lost half our solution space. The fix for complex roots is Euler's formula, already familiar: $e^{i\theta} = \cos\theta + i\sin\theta$. If $r=\alpha+i\beta$ solves the characteristic equation, then $e^{rt} = e^{\alpha t}(\cos\beta t + i \sin \beta t)$ solves the ODE as a complex-valued function. Because the ODE has real coefficients, the real and imaginary parts of any complex solution are themselves real solutions — a fact that converts one complex exponential into two real, independent, oscillating solutions. The picture is a spiral: $\alpha$ controls how fast the spiral expands or contracts (the envelope $e^{\alpha t}$), and $\beta$ controls how fast it winds around (the oscillation frequency). The fix for repeated roots is different in spirit: it is not an algebraic trick but a systematic method, reduction of order, that manufactures a second solution out of thin air once one solution is known. The intuition is that near a repeated root, the two nearby-but-distinct roots $r$ and $r+\epsilon$ of a perturbed problem give solutions $e^{rt}$ and $e^{(r+\epsilon)t}$; the combination $\frac{e^{(r+\epsilon)t}-e^{rt}}{\epsilon}$ stays finite as $\epsilon \to 0$ and limits to $te^{rt}$, which is exactly the "extra" solution needed to fill the two-dimensional gap.

Example 1 (complex roots, asymmetric coefficients). Solve the initial value problem

Characteristic equation: $2r^2-4r+10=0$, i.e. $r^2-2r+5=0$. Discriminant: $\Delta = (-2)^2-4(1)(5) = 4-20=-16<0$, confirming the complex case. Roots:

So $\alpha=1$, $\beta=2$. By the theorem, a fundamental set is $y_1=e^{t}\cos 2t$, $y_2=e^t\sin 2t$, and the general solution is

Applying initial conditions. At $t=0$: $y(0)=C_1 = 3$, so $C_1=3$. Differentiate: $y'(t) = e^t(C_1\cos2t+C_2\sin2t) + e^t(-2C_1\sin2t+2C_2\cos2t)$. At $t=0$:

With $C_1=3$: $3+2C_2=1 \Rightarrow C_2 = -1$. Solution:

Check: $y(0)=1\cdot(3-0)=3$ ✓. Compute $y'(t)$ directly: $y' = e^t(3\cos2t-\sin2t) + e^t(-6\sin2t-2\cos2t) = e^t(\cos2t - 7\sin2t)$; at $t=0$, $y'(0)=1$ ✓. One can also verify $2y''-4y'+10y=0$ symbolically by substituting $y=e^t(3\cos2t-\sin2t)$ and using that both $e^t\cos2t,e^t\sin2t$ individually solve the equation, so any real linear combination does too, by linearity of the ODE operator.

Example 2 (repeated roots, asymmetric coefficients). Solve

Note the initial condition is posed at $t=-1$, not $t=0$, to test the general solution honestly. Characteristic equation: $r^2+6r+9=0$, i.e. $(r+3)^2=0$. Discriminant $\Delta = 36-36=0$, confirming the repeated case, with repeated root $r=-3$. General solution (by the theorem): $y(t) = (C_1+C_2t)e^{-3t}$. Apply $y(-1)=2$:

Differentiate: $y'(t) = C_2e^{-3t} - 3(C_1+C_2t)e^{-3t} = e^{-3t}\big[(C_2-3C_1) - 3C_2t\big]$. At $t=-1$:

So $4C_2 - 3C_1 = -5e^{-3}$. Solve the linear system. From the first equation, $C_1 = C_2 + 2e^{-3}$. Substitute:

Then $C_1 = e^{-3}+2e^{-3} = 3e^{-3}$. Solution:

Check at $t=-1$: $y(-1) = (3-1)e^{0} = 2$ ✓. Compute $y'(t) = e^{-3(t+1)} - 3(3+t)e^{-3(t+1)} = e^{-3(t+1)}\big[1-3(3+t)\big] = e^{-3(t+1)}(-8-3t)$. At $t=-1$: $y'(-1) = e^0(-8+3) = -5$ ✓. This confirms the general repeated-root solution correctly absorbs an off-origin initial point.