Second-Order Homogeneous Equations

This is where physics truly lives. Newton's second law, $F = ma$, relates force to acceleration, which is the second derivative of position. Springs, pendulums, electrical circuits, vibrating strings: the equations that govern them all involve second derivatives. Understanding second-order differential equations unlocks the mathematics of motion and oscillation.

The Standard Form

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

where $a$, $b$, and $c$ are constants, and $a \neq 0$. The term homogeneous means the right side is zero: there is no external forcing function. The term linear means $y$, $y'$, and $y''$ appear only to the first power, with no products like $y \cdot y'$.

This seemingly simple form describes an enormous range of physical phenomena. A mass on a spring, a swinging pendulum with friction, the current in an RLC circuit: all are modeled by equations of exactly this type.

The Exponential Guess

From first-order equations, we know that $y = e^{kt}$ solves $y' = ky$. The exponential function has a remarkable property: its derivative is proportional to itself. This suggests trying $y = e^{rt}$ as a solution to our second-order equation.

If $y = e^{rt}$, then:

• $y' = re^{rt}$
• $y'' = r^2 e^{rt}$

Substituting into $ay'' + by' + cy = 0$:

Factor out $e^{rt}$, which is never zero:

Since $e^{rt} \neq 0$, we need:

This is the characteristic equation. The values of $r$ that satisfy it determine our solutions. What started as a differential equation has become a simple quadratic.

The Characteristic Equation

The characteristic equation $ar^2 + br + c = 0$ is solved using the quadratic formula:

The expression under the square root, $b^2 - 4ac$, is the discriminant. It determines the nature of the roots and, consequently, the character of the solutions.

Adjust the coefficients $a$, $b$, and $c$ to see how the roots move in the complex plane. When the discriminant is positive, both roots lie on the real axis. When negative, the roots become a complex conjugate pair. When the discriminant equals zero, the two roots coincide.

Case 1: Distinct Real Roots

When $b^2 - 4ac > 0$, we get two distinct real roots $r_1$ and $r_2$. Each gives a solution:

These solutions are linearly independent: neither is a constant multiple of the other. This means they span a two-dimensional solution space. The general solution is any linear combination:

The constants $C_1$ and $C_2$ are determined by initial conditions.

Watch how the solution is built from two exponential components. When both roots are negative, solutions decay to zero. When one root is positive, that component eventually dominates, causing the solution to grow without bound.

Case 2: Repeated Roots

When $b^2 - 4ac = 0$, the quadratic has a single repeated root:

Now we have a problem. The function $y_1 = e^{rt}$ is a solution, but where is the second independent solution? Simply using $e^{rt}$ twice does not give us two independent functions.

The answer is unexpected: the second solution is $y_2 = te^{rt}$.

Let us verify this. If $y = te^{rt}$, then:

• $y' = e^{rt} + rte^{rt} = e^{rt}(1 + rt)$
• $y'' = re^{rt}(1 + rt) + re^{rt} = re^{rt}(2 + rt)$

Let us verify this works. Substituting into $ay'' + by' + cy$:

Factoring out $e^{rt}$:

Since $r$ satisfies the characteristic equation, $ar^2 + br + c = 0$. And since $r = -b/(2a)$, we have $2ar + b = 2a(-b/(2a)) + b = 0$. Both terms vanish, confirming $te^{rt}$ is indeed a solution.

The general solution for repeated roots is:

The factor $te^{rt}$ starts at zero and initially grows linearly (when $r < 0$, it eventually decays as the exponential takes over). This polynomial-times-exponential form appears whenever roots repeat.

Why the $te^{rt}$ Solution?

There is an elegant way to understand where $te^{rt}$ comes from. Consider what happens as two distinct roots $r_1$ and $r_2$ approach each other.

The general solution is:

Suppose we want a solution with $y(0) = 0$ and $y'(0) = 1$. This gives $C_1 = -C_2 = 1/(r_1 - r_2)$, so:

Now let $r_1 \to r$ and $r_2 \to r$. This is a $0/0$ indeterminate form. By L'Hopital's rule (differentiating with respect to $r_1$):

The $te^{rt}$ solution emerges naturally as the limit when two independent exponential solutions merge into one.

Linear Independence and the General Solution

A second-order equation has a two-dimensional solution space. To span this space, we need two linearly independent solutions. Two functions $y_1$ and $y_2$ are linearly independent if the only constants satisfying $C_1 y_1 + C_2 y_2 = 0$ for all $t$ are $C_1 = C_2 = 0$.

For distinct real roots, $e^{r_1 t}$ and $e^{r_2 t}$ are independent because exponentials with different exponents cannot cancel each other.

For a repeated root, $e^{rt}$ and $te^{rt}$ are independent. If $C_1 e^{rt} + C_2 te^{rt} = 0$ for all $t$, then at $t = 0$ we get $C_1 = 0$, and then $C_2 te^{rt} = 0$ for all $t$ implies $C_2 = 0$.

This concept connects to linear algebra: the solutions form a vector space, and any two independent solutions form a basis for that space.

The Wronskian: A systematic way to test linear independence uses the Wronskian, defined for two functions $y_1$ and $y_2$ as:

Two solutions to a second-order equation are linearly independent if and only if their Wronskian is nonzero. For example, with $y_1 = e^{r_1 t}$ and $y_2 = e^{r_2 t}$:

The Wronskian will appear again when we study nonhomogeneous equations and variation of parameters.

Why Two Initial Conditions?

For a first-order equation, specifying $y(0)$ determines the solution uniquely. For a second-order equation, specifying $y(0)$ is not enough. We also need $y'(0)$.

The reason is geometric. Knowing the position of a particle tells you where it is, but not where it is going. Two particles at the same position but moving with different velocities will follow different trajectories.

The general solution has two free constants, $C_1$ and $C_2$. Two initial conditions provide two equations to solve for these constants. Specifying both $y(0)$ and $y'(0)$ picks out exactly one solution from the infinite family.

Knowing only the initial position leaves an entire family of solutions passing through that point. Adding the initial velocity narrows it down to exactly one.

Solving Initial Value Problems

Here is the procedure for solving $ay'' + by' + cy = 0$ with $y(0) = y_0$ and $y'(0) = v_0$:

1. Write the characteristic equation: $ar^2 + br + c = 0$
2. Find the roots using the quadratic formula
3. Write the general solution based on the root type
4. Apply initial conditions to find $C_1$ and $C_2$

Example: Solve $y'' + 3y' + 2y = 0$ with $y(0) = 1$ and $y'(0) = 0$.

The characteristic equation is $r^2 + 3r + 2 = 0$, which factors as $(r + 1)(r + 2) = 0$.

Roots: $r_1 = -1$, $r_2 = -2$ (distinct).

General solution: $y = C_1 e^{-t} + C_2 e^{-2t}$

Apply initial conditions:

• $y(0) = C_1 + C_2 = 1$
• $y'(0) = -C_1 - 2C_2 = 0$

Solving: $C_1 = 2$, $C_2 = -1$

Final answer: $y = 2e^{-t} - e^{-2t}$

Physical Interpretation

The sign of the roots determines long-term behavior:

• Both roots negative: Solutions decay to zero. This is overdamped motion, like a mass in thick honey returning to equilibrium.
• Both roots positive: Solutions grow without bound. This is unstable, like a pencil balanced on its tip.
• One positive, one negative: The positive root dominates at large $t$, causing growth.
• Repeated negative root: Solutions still decay, but the $te^{rt}$ term causes them to initially grow before decaying. This is critically damped motion.

The discriminant tells us whether the system is overdamped (distinct roots) or critically damped (repeated root). In the next chapter, we will see what happens when the discriminant is negative: the roots become complex, and solutions oscillate.

Looking Ahead

We have handled the cases where the characteristic equation has real roots. But what happens when $b^2 - 4ac < 0$? The quadratic formula gives complex numbers. In the next chapter, Complex Roots and Oscillations, we will see that complex roots produce sinusoidal oscillations, connecting second-order equations to waves, vibrations, and the rhythms of the physical world.

Key Takeaways

• Second-order homogeneous equations $ay'' + by' + cy = 0$ arise throughout physics, particularly in mechanics and circuits
• The characteristic equation $ar^2 + br + c = 0$ determines the form of solutions
• Distinct real roots $r_1 \neq r_2$ give $y = C_1 e^{r_1 t} + C_2 e^{r_2 t}$
• Repeated roots give $y = (C_1 + C_2 t)e^{rt}$, where $te^{rt}$ is the second independent solution
• Two initial conditions ($y(0)$ and $y'(0)$) are needed because the solution space is two-dimensional
• The discriminant $b^2 - 4ac$ classifies solutions: positive for distinct roots, zero for repeated roots, negative for complex roots (next chapter)