2.1 Homogeneous Second-Order Linear DEs with Constant Coefficients

A complete interactive guide — from theory to practice

What You'll Learn

By the end of this lesson, you will be able to:

Identify and classify second-order linear homogeneous ODEs with constant coefficients
Write and solve the characteristic equation
Handle all three cases: distinct real roots, repeated roots, and complex roots
Solve initial value problems with boundary conditions
Model physical systems like mass-spring systems and RLC circuits

What is a Homogeneous Second-Order ODE?

A homogeneous second-order linear differential equation with constant coefficients has the standard form:

a\,y'' + b\,y' + c\,y = 0

where $a$, $b$, and $c$ are constants (with $a \neq 0$), and $y''$ denotes the second derivative. The equation is homogeneous because the right side equals zero (no forcing term). This stands in contrast to non-homogeneous equations $ay'' + by' + cy = g(t)$ where $g(t) \neq 0$.

Key Requirement

The coefficients $a$, $b$, $c$ must be constants. If they depend on $x$ or $t$, you need different methods (like Euler equations or reduction of order). Always check this first!

Why Does This Matter?

Second-order ODEs with constant coefficients appear everywhere in engineering: mechanical vibrations (mass-spring systems), electrical circuits (LC, RL, RLC circuits), structural dynamics, control systems, and more. The method of characteristic equations is one of the most powerful and elegant tools in applied mathematics — it transforms a differential problem into an algebraic one.

“

Leonhard Euler discovered the characteristic equation method in the 1740s, revolutionizing how mathematicians solve differential equations. By recognizing that exponential functions $e^{rx}$ satisfy constant-coefficient equations, he reduced the problem of solving ODEs to solving algebraic equations — an insight of remarkable power and elegance.

— Leonhard Euler (1707–1783), master of both analysis and applied mathematics

مَنْ سَلَكَ طَرِيقًا يَلْتَمِسُ فِيهِ عِلْمًا سَهَّلَ اللَّهُ لَهُ بِهِ طَرِيقًا إِلَى الْجَنَّةِ

"Whoever travels a path in search of knowledge, Allah will make easy for him a path to Paradise."

— Prophet Muhammad (Peace Be Upon Him), Sahih Muslim 2699

Just as the characteristic equation reveals the hidden structure of differential equations, each step you take to understand mathematics brings you closer to deeper knowledge. The journey of learning is itself a blessing.

The Characteristic Equation Method

The Big Idea: Exponential Trial Solution

The key insight is to assume that the solution has the form $y = e^{rx}$ for some constant $r$ to be determined. Let's see what happens when we substitute this into $ay'' + by' + cy = 0$:

y' = r\,e^{rx}, \quad y'' = r^2\,e^{rx}

Substituting into the original equation:

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

Factor out $e^{rx}$ (which is never zero):

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

For this to hold for all $x$, we need the polynomial to vanish:

The Characteristic Equation

Given the ODE $ay'' + by' + cy = 0$, the characteristic equation is:

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

The roots of this quadratic determine the form of the general solution. Solve using the quadratic formula:

r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The discriminant $\Delta = b^2 - 4ac$ tells you which case you're in.

Three Cases Based on Roots

Case 1: Two Distinct Real Roots ($\Delta > 0$)

When $b^2 - 4ac > 0$, the characteristic equation has two different real roots $r_1 \neq r_2$.

y = C_1\,e^{r_1 x} + C_2\,e^{r_2 x}

Both $e^{r_1 x}$ and $e^{r_2 x}$ are solutions, and they are linearly independent (one is not a constant multiple of the other), so their linear combination gives the general solution.

Case 2: Repeated Root ($\Delta = 0$)

When $b^2 - 4ac = 0$, there is a single repeated root $r_1 = r_2 = r$.

y = (C_1 + C_2\,x)\,e^{rx}

If we only use $C_1\,e^{rx}$, we don't have enough constants to satisfy two initial conditions. The solution $x\,e^{rx}$ is linearly independent from $e^{rx}$ — this is why the $x$ factor is essential.

Why the Extra $x$?

When the root is repeated, $e^{rx}$ is a solution to the homogeneous equation. By the reduction of order principle, a second linearly independent solution is $x\,e^{rx}$. Multiplying by $x$ breaks the linear dependence and gives us the degree of freedom we need for two arbitrary constants.

Case 3: Complex Conjugate Roots ($\Delta < 0$)

When $b^2 - 4ac < 0$, the roots are complex conjugates: $r = \alpha \pm \beta i$ where

\alpha = -\frac{b}{2a}, \quad \beta = \frac{\sqrt{4ac - b^2}}{2a}

The exponential solutions $e^{(\alpha \pm \beta i)x}$ involve complex numbers, but we can extract real-valued solutions using Euler's formula:

e^{(\alpha + \beta i)x} = e^{\alpha x}(\cos(\beta x) + i\sin(\beta x))

Taking the real and imaginary parts gives two linearly independent real solutions:

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

This oscillatory form captures damped oscillations (if $\alpha < 0$) or growing oscillations (if $\alpha > 0$).

The Discriminant as Your Roadmap

Always compute $\Delta = b^2 - 4ac$ first:

$\Delta > 0$ → Two distinct real roots, exponential solutions
$\Delta = 0$ → One repeated root, exponential × polynomial solutions
$\Delta < 0$ → Complex roots, oscillatory + exponential envelope

Step-by-Step Solution Method

Follow these five steps to solve any homogeneous constant-coefficient second-order ODE:

Write in standard form. Make sure your equation is $ay'' + by' + cy = 0$ where the coefficient of $y''$ is $a$. If needed, divide through to make this true.
Write the characteristic equation. Replace $y''$ with $r^2$, $y'$ with $r$, and $y$ with $1$: $$ar^2 + br + c = 0$$
Solve for the roots. Use the quadratic formula to find $r$. Compute the discriminant $\Delta = b^2 - 4ac$ to determine which case applies.
Apply the case formula. Based on the roots, write the general solution:
- Distinct real $r_1, r_2$: $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}$
- Repeated root $r$: $y = (C_1 + C_2 x)e^{rx}$
- Complex $r = \alpha \pm \beta i$: $y = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))$
Apply initial conditions (if given). If you have $y(x_0) = y_0$ and $y'(x_0) = v_0$, substitute into the general solution and its derivative to get two equations for $C_1$ and $C_2$.

Shortcut: The Guessing Strategy

Once you master the three cases, you can often skip the full derivation in examples. Identify the equation, find the roots, and immediately write the solution form. This speed comes with practice!

Worked Examples

Click each example to reveal the full step-by-step solution.

Standard form: Already in form $y'' + y' - 6y = 0$ with $a=1, b=1, c=-6$.

Characteristic equation:

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

Solve for roots: Factor: $(r+3)(r-2) = 0$, so $r_1 = 2$ and $r_2 = -3$.

Discriminant check: $\Delta = 1^2 - 4(1)(-6) = 1 + 24 = 25 > 0$ ✓ Two distinct real roots.

Apply Case 1 formula:

\boxed{y(t) = C_1\,e^{2t} + C_2\,e^{-3t}}

Interpretation

The two terms represent two independent exponential modes. The $e^{2t}$ term grows, while $e^{-3t}$ decays. The general solution is their linear combination. For any initial conditions, we can find $C_1$ and $C_2$.

Standard form: $a=1, b=-4, c=4$.

Characteristic equation:

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

Solve for roots: Factor: $(r-2)^2 = 0$, so $r = 2$ (repeated).

Discriminant check: $\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$ ✓ Repeated root.

Apply Case 2 formula:

\boxed{y(t) = (C_1 + C_2\,t)e^{2t}}

Critical Point

If we only used $y = C_1 e^{2t}$, we'd have only one arbitrary constant — not enough for a second-order ODE. The $t$ factor in the second term $C_2\,t\,e^{2t}$ is essential. It's linearly independent from $e^{2t}$ because $t$ is not a constant.

Standard form: $a=1, b=0, c=9$.

Characteristic equation:

$$r^2 + 9 = 0$$

Solve for roots: $r^2 = -9$, so $r = \pm 3i$.

Discriminant check: $\Delta = 0^2 - 4(1)(9) = -36 < 0$ ✓ Complex roots.

Extract $\alpha$ and $\beta$: $r = 0 \pm 3i$, so $\alpha = 0$ and $\beta = 3$.

Apply Case 3 formula:

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

Physical Meaning

This is the equation for a mass on a spring with no damping. The solution oscillates indefinitely at frequency $\beta = 3$ rad/s. Real-world systems have some damping, which introduces the $e^{\alpha t}$ envelope to make oscillations decay.

Standard form: $a=1, b=-2, c=5$.

Characteristic equation:

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

Solve for roots: Using 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

Discriminant check: $\Delta = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0$ ✓ Complex roots.

Extract $\alpha$ and $\beta$: $\alpha = 1$, $\beta = 2$.

Apply Case 3 formula:

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

Damped vs. Underdamped Oscillation

Here $\alpha = 1 > 0$, so the oscillations grow in amplitude at rate $e^t$. This is an underdamped system with exponential instability. In a real mass-spring-dashpot system, negative damping (energy input) would create this behavior.

Standard form: $a=1, b=-6, c=8$.

Characteristic equation: $r^2 - 6r + 8 = 0$ → $(r-2)(r-4) = 0$ → $r_1 = 2, r_2 = 4$.

General solution:

y(x) = C_1\,e^{2x} + C_2\,e^{4x}

Find derivative:

y'(x) = 2C_1\,e^{2x} + 4C_2\,e^{4x}

Apply initial condition $y(0) = -2$:

$$-2 = C_1 + C_2$$

Apply initial condition $y'(0) = 6$:

$$6 = 2C_1 + 4C_2$$

Solve the system: From the first equation, $C_1 = -2 - C_2$. Substitute into the second:

$$6 = 2(-2 - C_2) + 4C_2 = -4 - 2C_2 + 4C_2 = -4 + 2C_2$$

10 = 2C_2 \quad \Rightarrow \quad C_2 = 5

Thus $C_1 = -2 - 5 = -7$.

Particular solution:

\boxed{y(x) = -7\,e^{2x} + 5\,e^{4x}}

Verification

$y(0) = -7(1) + 5(1) = -2$ ✓ and $y'(0) = -7(2) + 5(4) = -14 + 20 = 6$ ✓. Always verify your constants!

Standard form: $a=1, b=2, c=2$.

Characteristic equation: $r^2 + 2r + 2 = 0$.

Quadratic formula:

r = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i

Discriminant check: $\Delta = 2^2 - 4(1)(2) = 4 - 8 = -4 < 0$ ✓ Complex roots.

Extract $\alpha$ and $\beta$: $\alpha = -1$, $\beta = 1$.

General solution:

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

Physical Meaning

This is a damped oscillator with decay rate $\alpha = -1 < 0$. The oscillations occur at frequency $\beta = 1$ but decay exponentially as $e^{-t}$. Eventually, the system comes to rest. This model applies to real mechanical and electrical systems with energy dissipation.

Standard form: $a=1, b=-8, c=17$.

Characteristic equation: $r^2 - 8r + 17 = 0$.

Quadratic formula:

r = \frac{8 \pm \sqrt{64 - 68}}{2} = \frac{8 \pm \sqrt{-4}}{2} = \frac{8 \pm 2i}{2} = 4 \pm i

Extract $\alpha$ and $\beta$: $\alpha = 4$, $\beta = 1$.

General solution:

\boxed{y(t) = e^{4t}(C_1\cos(t) + C_2\sin(t))}

Unstable Oscillation

With $\alpha = 4 > 0$, the system exhibits growing oscillations. This could represent a feedback-driven system or a negative-damping scenario. The amplitude increases exponentially at rate $e^{4t}$.

Standard form: $a=1, b=0, c=-9$.

Characteristic equation: $r^2 - 9 = 0$.

Solve for roots: $r^2 = 9$ → $r = \pm 3$.

Discriminant check: $\Delta = 0^2 - 4(1)(-9) = 36 > 0$ ✓ Two distinct real roots.

General solution:

\boxed{y(t) = C_1\,e^{3t} + C_2\,e^{-3t}}

Alternative form (hyperbolic): Some prefer to write this using hyperbolic functions:

y(t) = A\cosh(3t) + B\sinh(3t)

where $A = C_1 + C_2$ and $B = C_1 - C_2$. Both forms are equivalent and useful in different contexts.

Hyperbolic vs. Exponential Form

For problems with symmetric real roots (like $\pm 3$), solutions are often written using hyperbolic functions $\cosh$ and $\sinh$ because they better represent "saddle-point" behavior. However, the exponential form is equally valid.

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.

Practice 1: Solve $y'' + 6y' + 9y = 0$ ▶

Solution

Step 1: Write the characteristic equation. For $y'' + 6y' + 9y = 0$, the characteristic equation is:

$$r^2 + 6r + 9 = 0$$

Step 2: Identify the discriminant. $\Delta = b^2 - 4ac = 36 - 36 = 0$. This indicates a repeated root.

Step 3: Factor the characteristic equation.

(r+3)^2 = 0 \Rightarrow r = -3 \text{ (repeated)}

Step 4: Apply the repeated root formula. When $r$ is repeated, the general solution is $y = (c_1 + c_2 t)e^{rt}$:

\boxed{y = (c_1 + c_2 t)e^{-3t}}

Note: The two linearly independent solutions are $e^{-3t}$ and $te^{-3t}$.

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

Solution

Step 1: Write the characteristic equation.

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

Step 2: Check the discriminant. $\Delta = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0$. This indicates complex roots.

Step 3: Use the quadratic formula.

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

Step 4: Identify $\alpha$ and $\beta$. From $r = 1 \pm 2i$, we have $\alpha = 1$ and $\beta = 2$.

Step 5: Apply the complex root formula. For $r = \alpha \pm i\beta$, the general solution is $y = e^{\alpha t}(c_1\cos(\beta t) + c_2\sin(\beta t))$:

\boxed{y = e^{t}(c_1\cos(2t) + c_2\sin(2t))}

Practice 3: IVP — Solve $y'' - 5y' + 6y = 0$, $y(0)=2$, $y'(0)=3$ ▶

Solution

Step 1: Write the characteristic equation.

$$r^2 - 5r + 6 = 0$$

Step 2: Factor. $(r-2)(r-3) = 0 \Rightarrow r = 2, 3$. Two distinct real roots.

Step 3: General solution.

y(t) = c_1 e^{2t} + c_2 e^{3t}

Step 4: Apply initial condition $y(0) = 2$.

$$2 = c_1 e^0 + c_2 e^0 = c_1 + c_2$$

Step 5: Find the derivative.

y'(t) = 2c_1 e^{2t} + 3c_2 e^{3t}

Step 6: Apply initial condition $y'(0) = 3$.

$$3 = 2c_1 + 3c_2$$

Step 7: Solve the system. From Step 4, $c_1 = 2 - c_2$. Substitute:

$$3 = 2(2 - c_2) + 3c_2 = 4 - 2c_2 + 3c_2 = 4 + c_2$$

c_2 = -1, \quad c_1 = 2 - (-1) = 3

Step 8: Particular solution.

\boxed{y(t) = 3e^{2t} - e^{3t}}

Interactive Visualizer

Enter coefficients $a$, $b$, $c$ for the equation $ay'' + by' + cy = 0$. The tool will compute the roots, show the discriminant, and plot the general solution for your choice of initial conditions.

Select Example

a (coeff of y'')

b (coeff of y')

c (coeff of y)

Characteristic Equation:
Discriminant:
Roots:
Solution Form:

C₁ (initial constant)

C₁ = 1.0

C₂ (initial constant)

C₂ = 1.0

x range

Practice Problems

Test your understanding! Try each problem, then check your answer.

Progress: 0 / 5 completed

Problem 1 — Characteristic Equation

What is the characteristic equation for $y'' + 5y' - 6y = 0$?

Problem 2 — Root Classification

What type of roots does the characteristic equation $r^2 - 4r + 4 = 0$ have?

Problem 3 — Distinct Real Roots

What is the general solution of $y'' + 4y = 0$?

Problem 4 — Complex Root Solution Form

If the characteristic equation has roots $r = 3 \pm 2i$, what is the solution form?

Problem 5 — Initial Value Problem

Solve $y'' + 4y' + 4y = 0$, $y(0) = 1$, $y'(0) = 0$. What is $y(x)$?