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Differential Equations Unit 4: Second-order linear equations

Second-order linear equations with constant coefficients are solved with the characteristic equation. The type of roots decides the form of the solution.

Covers: Homogeneous with constant coefficients, characteristic equation

Study guide

A second-order linear homogeneous equation with constant coefficients has the form ay″ + by′ + cy = 0. Guessing y = e^(rx) turns it into the characteristic equation ar² + br + c = 0, a quadratic in r. The roots determine the general solution, which always has two arbitrary constants.

If the roots are distinct and real (r₁ ≠ r₂), the solution is y = c₁e^(r₁x) + c₂e^(r₂x). If there's a repeated root r, it's y = (c₁ + c₂x)e^(rx); the extra factor of x provides a second independent solution. If the roots are complex, α ± βi, Euler's formula turns the exponentials into y = e^(αx)(c₁ cos βx + c₂ sin βx), an oscillation that grows, decays or stays steady depending on the sign of α.

The superposition principle says any combination of solutions to a linear homogeneous equation is also a solution. Two solutions form a fundamental set if they're linearly independent, which you can check with the Wronskian W = y₁y₂′ − y₂y₁′: if W ≠ 0, the general solution is c₁y₁ + c₂y₂.

An initial value problem gives y(x₀) and y′(x₀). Plug them into the general solution and its derivative to get two equations for c₁ and c₂. Boundary value problems instead give values at two different points and may have one, none or infinitely many solutions.

Euler's formula says e^(iθ) = cos θ + i sin θ. So e^((a+bi)t) = e^(at)(cos bt + i sin bt), and the real and imaginary parts of e^((α+βi)t) are the two real solutions e^(αt)cos βt and e^(αt)sin βt. The characteristic-equation method works for any order: an nth-order equation has a degree-n characteristic polynomial, and each root contributes solutions. A root r of multiplicity m contributes e^(rt), te^(rt), …, t^(m−1)e^(rt), and a repeated complex pair contributes the same powers of t times e^(αt)cos βt and e^(αt)sin βt.

An equation of the form ax²y″ + bxy′ + cy = 0 (for x > 0) is solved by guessing y = xʳ, which gives the indicial equation ar(r − 1) + br + c = 0. A repeated root r gives xʳ and xʳ ln x. Complex roots α ± βi give x^α cos(β ln x) and x^α sin(β ln x).

Key ideas

• ay″ + by′ + cy = 0 → characteristic equation ar² + br + c = 0.
• Two real roots r₁ ≠ r₂: y = c₁e^(r₁x) + c₂e^(r₂x).
• Repeated root r: y = (c₁ + c₂x)e^(rx).
• Complex roots α ± βi: y = e^(αx)(c₁ cos βx + c₂ sin βx).
• The Wronskian checks that two solutions are independent.

Formulas and rules

Common mistakes

• Forgetting the extra x for a repeated root.
• Only one arbitrary constant in a second-order solution.

Words to know

Characteristic equation of ay″ + by′ + cy = 0

ar² + br + c = 0

Solution with distinct real roots

y = c₁e^(r₁x) + c₂e^(r₂x)

Solution with a repeated root r

y = (c₁ + c₂x)e^(rx)

Solution with roots α ± βi

y = e^(αx)(c₁ cos βx + c₂ sin βx)

Second-order general solution

Has two arbitrary constants.

Second-order linear homogeneous equation

ay″ + by′ + cy = 0

Characteristic equation

ar² + br + c = 0

Distinct real roots solution

y = c₁e^(r₁x) + c₂e^(r₂x)

Repeated root solution

y = (c₁ + c₂x)e^(rx)

Complex roots solution

y = e^(αx)(c₁ cos βx + c₂ sin βx)

Superposition principle

Combinations of solutions of a linear homogeneous DE are solutions.

Wronskian

W = y₁y₂′ − y₂y₁′; W ≠ 0 means a fundamental set.

Fundamental set of solutions

Two independent solutions that generate the general solution.

Sample practice questions

23 more questions like these, from level 1 to 5, are in the practice set.