The Main Idea 

Second-order differential equations come in different types, and recognizing whether an equation is linear and whether it’s homogeneous determines which solution methods you can use. Understanding these classifications is the first step in solving differential equations.

Linear Second-Order Equations

A second-order differential equation is linear if it can be written as:

[latex]a_2(x)y'' + a_1(x)y' + a_0(x)y = r(x)[/latex]

The key requirement: [latex]y[/latex] and its derivatives appear only to the first power and are never multiplied together. Terms like [latex]y^2[/latex], [latex]\sqrt{y'}[/latex], [latex]\sin y[/latex], or [latex]yy'[/latex] make an equation nonlinear.

Homogeneous vs. Nonhomogeneous

Once you’ve confirmed an equation is linear, the next question is whether it’s homogeneous:

• Homogeneous: Every term contains [latex]y[/latex] or one of its derivatives. This means [latex]r(x) = 0[/latex] identically.
• Nonhomogeneous: At least one term involves only functions of [latex]x[/latex] (no [latex]y[/latex]). This means [latex]r(x) \neq 0[/latex] for some [latex]x[/latex].

Practical Strategy

When given a differential equation:

1. Rewrite in standard form if necessary
2. Check linearity: Are [latex]y[/latex] and its derivatives only to the first power? Are they never multiplied together or inside functions?
3. Check homogeneity: Does every term contain [latex]y[/latex] or its derivatives?

For example, [latex]xy'' + 2x^2y' + 5x^3y = 0[/latex] is linear and homogeneous, while [latex]xy'' + 2x^2y' + 5x^3y = x^2[/latex] is linear but nonhomogeneous (because of the [latex]x^2[/latex] term on the right).

Why This Matters

The classification determines your solution approach. Homogeneous linear equations have specific properties and solution methods that don’t apply to nonhomogeneous equations.

The Main Idea 

When solving second-order linear homogeneous differential equations, finding one solution isn’t enough. You need to find the general solution, which captures all possible solutions to the equation.

The Superposition Principle

If [latex]y_1(x)[/latex] and [latex]y_2(x)[/latex] are solutions to a linear homogeneous differential equation, then any linear combination [latex]y(x) = c_1y_1(x) + c_2y_2(x)[/latex] is also a solution. The constants [latex]c_1[/latex] and [latex]c_2[/latex] can be any real numbers.

Why Two Solutions Aren’t Always Enough

Here’s the catch: you can’t just pick any two solutions and combine them. For example, [latex]e^{-3t}[/latex] and [latex]2e^{-3t}[/latex] are both solutions to [latex]x'' + 7x' + 12x = 0[/latex], but their combination [latex]c_1e^{-3t} + c_2(2e^{-3t})[/latex] misses other solutions like [latex]e^{-4t}[/latex]. The problem is that these two solutions are essentially the same function—one is just a constant multiple of the other.

Linear Independence: The Key Requirement

Two functions are linearly independent if neither one is a constant multiple of the other (and neither is identically zero). For practical checking:

• [latex]x^2[/latex] and [latex]5x^2[/latex] are linearly dependent (one is 5 times the other)
• [latex]\sin x[/latex] and [latex]\cos x[/latex] are linearly independent (no constant [latex]C[/latex] makes [latex]\sin x = C\cos x[/latex])
• [latex]e^{3x}[/latex] and [latex]e^{-3x}[/latex] are linearly independent (different exponents mean they’re independent)

Problem-Solving Strategy

Once you have two linearly independent solutions [latex]y_1(x)[/latex] and [latex]y_2(x)[/latex], write the general solution as [latex]y(x) = c_1y_1(x) + c_2y_2(x)[/latex]. This formula represents every possible solution to the differential equation. This is powerful: finding two linearly independent solutions means you’ve found all solutions.

The Main Idea 

Second-order differential equations with constant coefficients have the form [latex]ay'' + by' + cy = 0[/latex], where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are constants. The solution strategy relies on finding the characteristic equation.

Why Exponential Functions?

We try solutions of the form [latex]y(x) = e^{\lambda x}[/latex] because exponential functions have a special property: their derivatives are constant multiples of themselves. This helps all terms cancel out when we substitute into the differential equation. Substituting [latex]y = e^{\lambda x}[/latex] into the equation gives us [latex]e^{\lambda x}(a\lambda^2 + b\lambda + c) = 0[/latex]. Since [latex]e^{\lambda x}[/latex] is never zero, we need [latex]a\lambda^2 + b\lambda + c = 0[/latex]. This is the characteristic equation.

Three Cases Based on the Roots

The characteristic equation is a quadratic, so it has three possible types of roots:

• Case 1: Distinct Real Roots ([latex]\lambda_1[/latex] and [latex]\lambda_2[/latex]) General solution: [latex]y(x) = c_1e^{\lambda_1 x} + c_2e^{\lambda_2 x}[/latex]
• Case 2: Repeated Real Root ([latex]\lambda[/latex]) General solution: [latex]y(x) = c_1e^{\lambda x} + c_2xe^{\lambda x}[/latex] Note the [latex]x[/latex] multiplier in the second term—this ensures linear independence.
• Case 3: Complex Conjugate Roots ([latex]\alpha \pm \beta i[/latex]) General solution: [latex]y(x) = e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)[/latex] The real part [latex]\alpha[/latex] appears in the exponential, while the imaginary part [latex]\beta[/latex] appears in the trigonometric functions.

Problem-Solving Strategy

1. Write the equation in standard form: [latex]ay'' + by' + cy = 0[/latex]
2. Form the characteristic equation: [latex]a\lambda^2 + b\lambda + c = 0[/latex]
3. Solve for [latex]\lambda[/latex] using factoring or the quadratic formula
4. Use the type of roots to determine which case applies and write the general solution

The Main Idea 

Finding the general solution to a differential equation gives you a family of solutions with arbitrary constants. To find a specific solution that describes a particular physical system, you need additional information about the system at specific points in time.

Initial-Value Problems

An initial-value problem provides conditions at a single point in time [latex]t_0[/latex]. For second-order equations, you need two initial conditions: typically [latex]y(t_0)[/latex] and [latex]y'(t_0)[/latex]. These two conditions let you solve for the two arbitrary constants [latex]c_1[/latex] and [latex]c_2[/latex] in the general solution. Initial-value problems for constant-coefficient second-order equations always have a unique solution when two initial conditions are provided.

Boundary-Value Problems

A boundary-value problem provides conditions at two different points in time: [latex]y(t_0)[/latex] and [latex]y(t_1)[/latex]. Unlike initial-value problems, boundary-value problems are less predictable. Even with two boundary conditions, you might find a unique solution, infinitely many solutions, or no solution at all.

Problem-Solving Strategy

1. Find the general solution to the differential equation
2. Calculate [latex]y'(t)[/latex] from the general solution (if working with initial conditions involving [latex]y'(t_0)[/latex])
3. Substitute the given conditions into [latex]y(t)[/latex] and [latex]y'(t)[/latex] to create a system of equations
4. Solve the system for [latex]c_1[/latex] and [latex]c_2[/latex]
5. Write the specific solution by substituting the values of [latex]c_1[/latex] and [latex]c_2[/latex] back into the general solution

For boundary-value problems, watch for inconsistent equations (no solution) or dependent equations (infinitely many solutions).