Homogeneous Equations with Constant Coefficients

The characteristic equation method for $ay''+by'+cy=0$ is the entry point to an enormous range of applications. In mechanics, it governs the free (unforced) motion of a damped spring–mass system, $m\ddot{x} + \gamma\dot{x} + kx = 0$; distinct real roots correspond to the overdamped regime in which the mass returns to equilibrium without oscillating. In electrical engineering, the same equation with $a=L$ (inductance), $b=R$ (resistance), $c=1/C$ (inverse capacitance) describes the free response of a series RLC circuit, and overdamped roots describe circuits that decay to zero current without ringing. In control theory, the roots of the characteristic polynomial are precisely the poles of the system's transfer function, and their sign and reality determine stability and the qualitative shape of the response — real negative roots mean stable, non-oscillatory decay. The same idea, viewed through linear algebra, is the $2\times2$ special case of finding eigenvalues of the companion matrix associated to a linear ODE system; this connection generalizes the method to arbitrary order $n$ and to systems of first-order equations, a viewpoint essential in numerical methods and in the study of dynamical systems more broadly (Strogatz-style phase plane analysis of linear systems near an equilibrium relies on exactly this eigenvalue classification). In probability and statistics, linear recurrence relations with constant coefficients — the discrete analogue — are solved by an entirely parallel characteristic-root method, underlying the analysis of Markov chains and generating functions. In numerical analysis and scientific computing, understanding the exact solution behavior for distinct real roots provides the baseline against which numerical ODE solvers (Euler's method, Runge–Kutta) are validated for accuracy and stability.

Ask the tutor: How does the general solution formula change when the characteristic equation has complex conjugate roots instead of real distinct ones?

The exponential function $e^{rt}$ has a defining property that makes it the natural candidate for solving any linear equation with constant coefficients: differentiating it just multiplies it by $r$. So if $y = e^{rt}$, then $y' = re^{rt}$ and $y'' = r^2 e^{rt}$, and every term in $ay''+by'+cy$ becomes a multiple of the same function $e^{rt}$. The differential equation collapses to an algebraic statement about the exponent $r$: the polynomial $ar^2+br+c$ must vanish. This is the entire idea. Solving a second-order linear ODE with constant coefficients is equivalent to factoring a quadratic. Once we have two distinct real roots $r_1 \ne r_2$, we get two solution functions $e^{r_1t}$ and $e^{r_2t}$. Because the ODE is linear (a fact already used in constructing general solutions from fundamental sets), any combination $C_1 e^{r_1 t} + C_2 e^{r_2 t}$ is again a solution. The remaining question — whether this two-parameter family captures every solution — is answered by checking that $\{e^{r_1t}, e^{r_2t}\}$ is a fundamental set, i.e. that its Wronskian never vanishes. Geometrically, distinct real roots correspond to two different exponential growth/decay rates; the solution space is the plane spanned by these two "basis" curves, and no scalar combination of one can reproduce the other unless the rates coincide.

Example 1 — asymmetric coefficients, initial value problem. Solve $2y'' - y' - 6y = 0$ with $y(0) = 4$, $y'(0) = -1$. Step 1 — characteristic equation. Substituting $y=e^{rt}$ gives

The discriminant is $\Delta = (-1)^2 - 4(2)(-6) = 1 + 48 = 49 > 0$, so distinct real roots exist. By the quadratic formula,

Since $49$ is a perfect square this factors cleanly: $2r^2-r-6 = (r-2)(2r+3)$, confirming $r_1=2, r_2=-3/2$. Step 2 — general solution. By the theorem,

Step 3 — apply initial conditions. Compute $y'(t) = 2C_1e^{2t} - \tfrac{3}{2}C_2 e^{-3t/2}$. At $t=0$:

From the first equation, $C_2 = 4 - C_1$. Substitute:

Then $C_2 = 4 - \tfrac{10}{7} = \tfrac{18}{7}$. Step 4 — check. At $t=0$: $y(0) = \tfrac{10}{7}+\tfrac{18}{7} = \tfrac{28}{7}=4$ ✓. And $y'(0) = 2\cdot\tfrac{10}{7} - \tfrac32\cdot\tfrac{18}{7} = \tfrac{20}{7} - \tfrac{27}{7} = -\tfrac{7}{7} = -1$ ✓. So

This solution grows without bound as $t\to\infty$ (dominated by the $e^{2t}$ term) and tends to $+\infty$ as $t \to -\infty$ too (dominated there by $e^{-3t/2}$), illustrating a saddle-type behavior typical of roots of opposite sign.

Example 2 — both roots negative, boundary-style data at a nonzero base point. Solve $y'' + 5y' + 6y = 0$ subject to $y(1) = 2$, $y'(1) = 0$. Step 1 — characteristic equation. Here $a=1, b=5, c=6$:

Discriminant check: $\Delta = 25 - 24 = 1 > 0$, consistent with two distinct real roots. Step 2 — general solution.

Step 3 — apply data at $t_0 = 1$ (not $0$, to test the general argument in Step 4 of the theorem). We have $y'(t) = -2C_1e^{-2t} - 3C_2e^{-3t}$, so

Let $u = C_1 e^{-2}$ and $v = C_2 e^{-3}$ to simplify: $u+v=2$ and $-2u-3v=0$, i.e. $2u = -3v$, i.e. $u = -\tfrac32 v$. Substitute into the first equation:

Then $u = -\tfrac32(-4) = 6$. Converting back,

Step 4 — check. $y(1) = 6e^2\cdot e^{-2} + (-4e^3)\cdot e^{-3} = 6 - 4 = 2$ ✓. And $y'(1) = -2(6e^2)e^{-2} -3(-4e^3)e^{-3} = -12 + 12 = 0$ ✓. So

Both exponents decay ($r_1,r_2<0$), so $y(t)\to 0$ as $t\to\infty$: this is the overdamped-decay regime, exactly matching a mechanical system with heavy damping and no oscillation.