Second-Order Linear ODE with Constant Coefficients

The general form

A homogeneous second-order linear ODE with constant coefficients has the form $a \, y'' + b \, y' + c \, y = 0, \qquad a, b, c \in \mathbb{R}, \; a \ne 0.$

"Homogeneous" here means the right side is $0$ (the #178 sense doesn't apply — different usage). "Constant coefficients" means $a, b, c$ don't depend on $x$.

The whole method to solve these is one slick idea: guess $y = e^{r x}$. Because plugging that in gives $a r^{2} e^{r x} + b r e^{r x} + c e^{r x} = e^{r x} (a r^{2} + b r + c) = 0,$ and $e^{r x} \ne 0$, the equation collapses to the algebraic characteristic equation $\boxed{\, a r^{2} + b r + c = 0 \,}$

Solve this quadratic for $r$ — three possible outcomes, three kinds of solutions. (Repeated roots are in #185; complex roots are in #184.) This problem focuses on the clean case of two real distinct roots.

The given equation

$y'' - 5 y' + 6 y = 0$

Coefficients $a = 1$, $b = -5$, $c = 6$.

Step-by-step solution

Step 1 — Write the characteristic equation. $r^{2} - 5 r + 6 = 0$

Step 2 — Solve it.

Factor: $r^{2} - 5 r + 6 = (r - 2)(r - 3) = 0$. $r_1 = 2, \quad r_2 = 3$

(Or use the quadratic formula: $r = (5 \pm \sqrt{25 - 24})/2 = (5 \pm 1)/2$.)

Step 3 — Write the two basic solutions. $y_1 = e^{2 x}, \qquad y_2 = e^{3 x}$

Step 4 — Combine into the general solution.

Because the ODE is linear and homogeneous, any linear combination of solutions is also a solution (see #200 on superposition). The general solution is $\boxed{\, y(x) = C_1 \, e^{2 x} + C_2 \, e^{3 x} \,}$

for arbitrary constants $C_1, C_2$.

Verification

Differentiate $y = C_1 e^{2x} + C_2 e^{3x}$: $y' = 2 C_1 e^{2x} + 3 C_2 e^{3x}$ $y'' = 4 C_1 e^{2x} + 9 C_2 e^{3x}$

Plug into the ODE $y'' - 5 y' + 6 y$: $4 C_1 e^{2x} + 9 C_2 e^{3x} - 5(2 C_1 e^{2x} + 3 C_2 e^{3x}) + 6 (C_1 e^{2x} + C_2 e^{3x})$ $= C_1 e^{2x}(4 - 10 + 6) + C_2 e^{3x}(9 - 15 + 6) = 0 + 0 = 0 \quad\checkmark$

The clean cancellation comes from the fact that each $r$ satisfies $r^2 - 5r + 6 = 0$.

Why two arbitrary constants?

The ODE is order 2, so its general solution has a 2-parameter family. Geometrically, the solution space is a 2-dimensional vector space (over the reals), spanned by the basis $\{e^{2x}, e^{3x}\}$.

Two pieces of data (typically an IVP $y(x_0) = a$, $y'(x_0) = b$, or a BVP with boundary values at two endpoints) pin down $C_1$ and $C_2$.

Initial value problem

Say $y(0) = 0$ and $y'(0) = 1$. Then $0 = C_1 + C_2$ $1 = 2 C_1 + 3 C_2$

From the first, $C_2 = -C_1$. Substitute: $1 = 2 C_1 - 3 C_1 = -C_1 \implies C_1 = -1$, $C_2 = 1$. $y(x) = -e^{2 x} + e^{3 x} = e^{2 x} (e^{x} - 1).$

Check: $y(0) = 0$ ✓; $y'(x) = -2 e^{2x} + 3 e^{3x}$, $y'(0) = -2 + 3 = 1$ ✓.

The three cases of the characteristic equation

Let the discriminant be $\Delta = b^{2} - 4 a c$. Then:

• $\Delta > 0$: two real distinct roots $r_1, r_2$. General solution $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}$. (This problem. Also #183.)
• $\Delta = 0$: repeated real root $r$. General solution $y = (C_1 + C_2 x) e^{r x}$. (See #185.)
• $\Delta < 0$: complex conjugate roots $\alpha \pm i \beta$. General solution $y = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x)$. (See #184.)

Pattern: you always get exactly $\dim = 2$ linearly independent solutions, but their form depends on which case the discriminant falls into.

Stability at a glance

Just by looking at the roots you can predict long-term behaviour:

• Both $r < 0$: solutions decay to $0$ — stable.
• Both $r > 0$: solutions grow unboundedly — unstable.
• One $r > 0$, one $r < 0$: saddle — generic initial data blows up, but a 1-parameter family of initial data decays.

For our problem $r = 2, 3$ — both positive, so every non-trivial solution grows exponentially. No equilibrium other than $y \equiv 0$.

Common mistakes

• Forgetting to divide by $a$. If $a \ne 1$, the characteristic equation is $r^{2} + (b/a) r + (c/a) = 0$, not $r^{2} + b r + c$. (Or keep $a$ and solve $a r^{2} + b r + c = 0$ directly — either way, don't drop $a$.)
• Using only one constant. A second-order linear ODE needs two arbitrary constants in its general solution.
• Plugging in before solving for $C$. Apply initial conditions after you have the general solution with both constants.
• Treating non-distinct roots the same way. If $r_1 = r_2$, then $e^{r_1 x}$ and $e^{r_2 x}$ are the same solution — you need to manufacture a second linearly independent one (it's $x \, e^{r x}$, see #185).

Try it in the visualization

Slide the coefficients $b$ and $c$ and watch the discriminant cross zero. See the solution morph from two exponentials (Δ>0) into a repeated-root form (Δ=0) into oscillating exponentials (Δ<0). Each regime has a distinctly different shape.