Chapter 2 · Second-Order ODEs

Homogeneous Equations with Constant Coefficients

The Characteristic Equation & Three Cases

Dr. Mohamed Mabrok · Qatar University

The Big Picture: Second-Order Linear ODEs

Homogeneous
(g = 0)

This Lecture!

ay'' + by' + cy = g(t)

Non-Homogeneous
(g ≠ 0)

Undetermined Coefficients

Variation of Parameters

Variable Coefficients

Euler Equations

Why Second-Order Equations Matter

⚙️

Mass-Spring Systems

Mechanical vibrations, oscillations, and damping.

$$my'' + cy' + ky = 0$$

⚡

RLC Circuits

Electrical oscillations, resonance, and energy dissipation.

LQ'' + RQ' + \frac{Q}{C} = 0

🏗️

Beam Deflection

Structural engineering, bridge design, safety analysis.

$$EIy'' = M(x)$$

Real-World Impact

      Second-order ODEs are everywhere in engineering: vehicle suspension systems, earthquake-resistant buildings, aircraft wing design, and more.

Standard Form & The Brilliant Idea

Standard Homogeneous Form

      $$ay'' + by' + cy = 0$$
    
      where $a \neq 0$, and $a, b, c$ are constants.

The Key Insight

      What if we try $y = e^{rx}$ for some constant $r$?
    
      y = e^{rx} \quad \Rightarrow \quad y' = re^{rx} \quad \Rightarrow \quad y'' = r^2e^{rx}

SUBSTITUTE into the ODE

a(r^2e^{rx}) + b(re^{rx}) + c(e^{rx}) = 0

e^{rx}(ar^2 + br + c) = 0

The Characteristic Equation!

      $$ar^2 + br + c = 0$$

The Three Cases: Solving the Characteristic Equation

The discriminant $\Delta = b^2 - 4ac$ determines the nature of roots.

Case 1: Distinct Real Roots

Δ > 0

$r_1 \neq r_2$ (both real)

y = C_1e^{r_1x} + C_2e^{r_2x}

Case 2: Repeated Root

Δ = 0

$r_1 = r_2 = r$ (double root)

y = (C_1 + C_2x)e^{rx}

Case 3: Complex Roots

Δ < 0

$r = \alpha \pm \beta i$ (conjugates)

y = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)

Key Point

      Always compute the discriminant first! It tells you which case you're dealing with and which formula to use.

Case 1: Distinct Real Roots (Δ > 0)

General Solution

      $$y = C_1e^{r_1x} + C_2e^{r_2x}$$
    
      where $r_1 \neq r_2$ are the two distinct real roots of $ar^2 + br + c = 0$.

Why This Works

      Both $e^{r_1x}$ and $e^{r_2x}$ are solutions. They are linearly independent (one is not a constant multiple of the other) because $r_1 \neq r_2$. Any linear combination is also a solution!

Wronskian Check

      $$W = e^{r_1x} \cdot r_2e^{r_2x} - e^{r_2x} \cdot r_1e^{r_1x} = (r_2 - r_1)e^{(r_1+r_2)x} \neq 0$$
    
      Since $r_1 \neq r_2$, the Wronskian is never zero. This guarantees linear independence and that we have the general solution!

Worked Example

Solve: $y'' + y' - 6y = 0$

STEP 1 Write the characteristic equation

$$r^2 + r - 6 = 0$$

STEP 2 Factor the characteristic equation

$$(r + 3)(r - 2) = 0$$

STEP 3 Find the roots

r_1 = -3, \quad r_2 = 2

STEP 4 Write the general solution

y = C_1e^{-3t} + C_2e^{2t}

Case 2: Repeated Root (Δ = 0)

When $\Delta = 0$, the characteristic equation has a single double root $r$.

The Problem

      If $r$ is a double root, we only get one solution: $y_1 = e^{rx}$. But we need two linearly independent solutions to form the general solution!

The Solution

      Multiply the first solution by $x$! This gives us a second linearly independent solution:
    
      $$y_1 = e^{rx}, \quad y_2 = xe^{rx}$$

General Solution (Repeated Root)

      $$y = (C_1 + C_2x)e^{rx}$$
    
      where $r$ is the double root of $ar^2 + br + c = 0$.

Worked Example

Solve: $y'' - 4y' + 4y = 0$

STEP 1 Write the characteristic equation

$$r^2 - 4r + 4 = 0$$

STEP 2 Factor — this is a perfect square!

$$(r - 2)^2 = 0$$

STEP 3 Find the double root

r = 2 \text{ (double root)}

STEP 4 Write the general solution (with $x$ factor!)

y = (C_1 + C_2t)e^{2t}

Case 3: Complex Conjugate Roots (Δ < 0)

Complex Roots

      $$r = \alpha \pm \beta i$$
    
      where $\alpha = -\frac{b}{2a}$ and $\beta = \frac{\sqrt{4ac - b^2}}{2a}$

Euler's Formula (The Magic!)

      $$e^{(\alpha + \beta i)x} = e^{\alpha x}(\cos(\beta x) + i\sin(\beta x))$$
    
      Complex exponentials are really oscillations wrapped in an exponential envelope!

General Solution (Complex Roots)

      $$y = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))$$
    
      The solution oscillates (due to $\cos$ and $\sin$) with exponential growth/decay (due to $e^{\alpha x}$).

Worked Example

Solve: $y'' + 9y = 0$

STEP 1 Write the characteristic equation

$$r^2 + 9 = 0$$

STEP 2 Solve for $r$

r^2 = -9 \quad \Rightarrow \quad r = \pm 3i

STEP 3 Identify $\alpha$ and $\beta$

\alpha = 0, \quad \beta = 3

STEP 4 Write the general solution

y = C_1\cos(3t) + C_2\sin(3t)

Physical Meaning

      Pure oscillation with no decay! The envelope $e^{0 \cdot t} = 1$ means the oscillation continues forever at the same amplitude.

Worked Example

Solve: $y'' - 2y' + 5y = 0$

STEP 1 Write the characteristic equation

$$r^2 - 2r + 5 = 0$$

STEP 2 Use the quadratic formula

r = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i

STEP 3 Identify $\alpha$ and $\beta$

\alpha = 1, \quad \beta = 2

STEP 4 Write the general solution

y = e^{t}(C_1\cos(2t) + C_2\sin(2t))

Physical Meaning

      Growing oscillation! The oscillation (from $\cos$ and $\sin$) gets larger over time due to $e^{t}$ growing exponentially.

Worked Example

Solve the IVP: $y'' - 6y' + 8y = 0$, $y(0) = -2$, $y'(0) = 6$

STEP 1-2 Find general solution from characteristic equation

r^2 - 6r + 8 = 0 \quad \Rightarrow \quad (r-2)(r-4) = 0 \quad \Rightarrow \quad r_1 = 2, r_2 = 4

STEP 3 General solution

y = C_1e^{2x} + C_2e^{4x}

STEP 4 Apply $y(0) = -2$

-2 = C_1e^{0} + C_2e^{0} = C_1 + C_2

STEP 5 Apply $y'(0) = 6$ (compute $y'$ first)

y' = 2C_1e^{2x} + 4C_2e^{4x} \quad \Rightarrow \quad 6 = 2C_1 + 4C_2

STEP 6 Solve the system

C_1 + C_2 = -2 \quad \text{and} \quad 2C_1 + 4C_2 = 6 \quad \Rightarrow \quad C_1 = -7, C_2 = 5

y = -7e^{2x} + 5e^{4x}

What Do the Cases Mean Physically?

In a mass-spring-damper system: $my'' + cy' + ky = 0$, the three cases correspond to different damping regimes.

Overdamped

Two negative distinct roots
(Case 1: $r_1 < r_2 < 0$)

Damping is very high. The mass slowly returns to equilibrium without oscillating.

Critically Damped

One negative double root
(Case 2: $r < 0$)

Perfect balance. The mass returns to equilibrium as fast as possible without oscillating.

Underdamped

Complex roots with negative real part
(Case 3: $\alpha < 0$, $\beta \neq 0$)

Damping is weak. The mass oscillates around equilibrium while gradually decaying.

Intuition

      All three cases are stable (return to equilibrium) because the real parts of the roots are negative. The type of return depends on whether roots are real or complex!

Key Takeaways & Summary

Case 1: $\Delta > 0$

Two distinct real roots $r_1, r_2$

Solution: $y = C_1e^{r_1x} + C_2e^{r_2x}$

Case 2: $\Delta = 0$

One double root $r$

Solution: $y = (C_1 + C_2x)e^{rx}$

Case 3: $\Delta < 0$

Complex roots $\alpha \pm \beta i$

Solution: $y = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

⚠️ Common Mistakes to Avoid

Skipping the discriminant: Always check $\Delta = b^2 - 4ac$ first!
Forgetting the $x$ factor in Case 2: Repeated roots need $(C_1 + C_2x)e^{rx}$, not just $C_1e^{rx}$.
Confusing $\alpha$ and $\beta$ in Case 3: $\alpha$ is the real part (exponential envelope), $\beta$ is the imaginary part (oscillation frequency).
Forgetting to find constants from initial conditions: The general solution has arbitrary constants — use ICs to pin them down!

Next Up

      What happens when the right side is not zero ($g(t) \neq 0$)? That's non-homogeneous second-order ODEs — solved using Undetermined Coefficients and Variation of Parameters!