Second-Order ODE with Real Distinct Roots

Setup

The ODE $y'' - y = 0$ is the simplest "two real roots, opposite signs" example. It also happens to be the defining equation of the hyperbolic functions $\cosh$ and $\sinh$, so there's more structure here than meets the eye.

Step-by-step solution

Step 1 — Characteristic equation. $r^{2} - 1 = 0 \implies (r - 1)(r + 1) = 0 \implies r_1 = 1, \; r_2 = -1$

Discriminant $\Delta = 0 - 4(1)(-1) = 4 > 0$, two real distinct roots.

Step 2 — General solution. $\boxed{\, y(x) = C_1 \, e^{x} + C_2 \, e^{-x} \,}$

Verification

$y = C_1 e^{x} + C_2 e^{-x}$ $y' = C_1 e^{x} - C_2 e^{-x}$ $y'' = C_1 e^{x} + C_2 e^{-x}$

So $y'' - y = (C_1 e^{x} + C_2 e^{-x}) - (C_1 e^{x} + C_2 e^{-x}) = 0$. ✓

The two exponential modes

Any solution decomposes uniquely into two modes:

• $y_1 = e^{x}$: the growing mode. $r_1 = 1 > 0$ drives exponential growth.
• $y_2 = e^{-x}$: the decaying mode. $r_2 = -1 < 0$ drives exponential decay.

Every non-trivial solution eventually blows up (because the growing mode dominates) unless $C_1 = 0$, in which case the solution decays to $0$. Setting $C_1 = 0$ requires precisely the right initial conditions — it's the stable manifold of the saddle at the origin. This is the defining feature of a "saddle" fixed point: one direction attracts, the other repels.

Recombination into hyperbolic functions

$e^{x}$ and $e^{-x}$ are one basis for the 2-d solution space. So are their sum and difference (halved): $\cosh x = \frac{e^{x} + e^{-x}}{2}, \qquad \sinh x = \frac{e^{x} - e^{-x}}{2}.$

So we can equivalently write $y(x) = A \cosh x + B \sinh x$ with $A, B$ a linear rebasing of $C_1, C_2$ ($A = C_1 + C_2$, $B = C_1 - C_2$).

When to prefer which basis?

• Exponentials are natural for IVPs with boundary at $\pm \infty$ (growing / decaying behaviour explicit).
• Hyperbolics are natural for symmetric BVPs, e.g. $y(-L) = y(L) = 1$ (even solution → pick $B = 0$).

Our ODE's solution space is one two-dimensional vector space; $\{e^{x}, e^{-x}\}$ and $\{\cosh x, \sinh x\}$ are just two different bases.

Initial value problem

$y(0) = 2$, $y'(0) = 0$.

Using the hyperbolic basis: $y(0) = A \cosh 0 + B \sinh 0 = A = 2$ $y'(0) = A \sinh 0 + B \cosh 0 = B = 0$ $y(x) = 2 \cosh x$

Using the exponential basis (equivalent): $2 = C_1 + C_2, \quad 0 = C_1 - C_2 \implies C_1 = C_2 = 1$ $y(x) = e^{x} + e^{-x} = 2 \cosh x \; \checkmark$

Geometric picture — catenary and saddle

Catenary (hanging chain) is shaped like $y = a \cosh(x/a)$ — solutions of $y'' = y / a^2$ with appropriate scaling. The mathematics of hanging cables, power lines, and arches all live in this one ODE.

In the phase plane $(y, y')$ the ODE $y'' = y$ becomes the 2D linear system $\dot y = v, \quad \dot v = y.$ Eigenvalues are $\pm 1$ — one positive, one negative — the classic saddle. Most initial conditions eventually go to $\infty$; only the 1-d stable manifold (the direction of $e^{-x}$ mode) gets pulled in to $0$.

Contrast with $y'' + y = 0$ (the harmonic oscillator)

A sign flip gives $y'' + y = 0$, characteristic equation $r^{2} + 1 = 0$, roots $r = \pm i$. Pure imaginary roots ⇒ solutions are $\cos x, \sin x$ (bounded oscillation, see #184). Same machinery, but the sign of the discriminant flips us from exponentials-that-blow-up to pure oscillations.

Common mistakes

• Writing $y = C_1 e^{x} - C_2 e^{-x}$ with a minus sign glued to $C_2$. The constants are arbitrary — you don't need a sign in front, just absorb it into $C_2$.
• Thinking exponentials and hyperbolics are different solutions. They span the same space; it's just a change of basis.
• Missing the stable/unstable split. For any saddle-type ODE there's a 1-d stable manifold of initial data that decays — worth finding explicitly.

Try it in the visualization

Slide $C_1, C_2$ independently and watch the two exponential modes add up. Toggle the basis to $\{\cosh, \sinh\}$ to see the same solution written differently — especially useful for symmetric initial data.