Second-Order Differential Equations: Auxiliary Equation Method

By Engr. Henry Akpan

Equations of Vibration

A mass bouncing on a spring, a swinging pendulum, the current in a radio tuning circuit and the sway of a building in the wind all obey the same kind of equation: one involving a second derivative. These are second-order differential equations, and the most common type, the linear equation with constant coefficients, can be solved by a neat, systematic method. This tutorial, for MTH 201/202 and engineering mathematics, covers the auxiliary (characteristic) equation and its three cases, the general solution, initial conditions, non-homogeneous equations by the method of undetermined coefficients, and applications to simple and damped harmonic motion. It follows on from the tutorial on first-order differential equations.

The Standard Form

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

a y″ + b y′ + c y = f(x)

Here y′ = dy/dx, y″ = d²y/dx², and a, b and c are constants.

  • If f(x) = 0, the equation is homogeneous.
  • Otherwise it is non-homogeneous, and f(x) is often called the forcing term.

The general solution of a second-order equation contains two arbitrary constants, because solving it in effect involves two integrations.

Solving the Homogeneous Equation

Try a solution of the form y = e^(mx). Then y′ = m e^(mx) and y″ = m² e^(mx). Substituting into a y″ + b y′ + c y = 0 and dividing by e^(mx), which is never zero, gives the auxiliary equation:

am² + bm + c = 0

This is an ordinary quadratic. The form of the solution depends on the type of its roots.

Case 1: two distinct real roots, m₁ and m₂

y = A e^(m₁x) + B e^(m₂x)

Worked example 1

Solve y″ − 5y′ + 6y = 0.

The auxiliary equation is m² − 5m + 6 = 0, so (m − 2)(m − 3) = 0, and m = 2 or m = 3.

General solution: y = A e^(2x) + B e^(3x).

Case 2: a repeated real root, m

One solution is e^(mx), and the second independent solution turns out to be x e^(mx):

y = (A + Bx) e^(mx)

Worked example 2

Solve y″ − 4y′ + 4y = 0.

m² − 4m + 4 = 0, so (m − 2)² = 0, and m = 2 is a repeated root.

General solution: y = (A + Bx) e^(2x).

Case 3: complex roots, m = α ± βi

Using Euler's formula, the complex exponentials combine into real sines and cosines:

y = e^(αx)(A cos βx + B sin βx)

Worked example 3

Solve y″ + 2y′ + 5y = 0.

m² + 2m + 5 = 0, so m = (−2 ± √(4 − 20)) ÷ 2 = −1 ± 2i. So α = −1 and β = 2.

General solution: y = e^(−x)(A cos 2x + B sin 2x), which is an oscillation that dies away.

Initial Conditions

Two conditions, usually the values of y and y′ at a starting point, fix the two constants and give a particular solution.

Worked example 4

Solve y″ − 5y′ + 6y = 0, given that y(0) = 1 and y′(0) = 0.

From example 1, y = A e^(2x) + B e^(3x), so y′ = 2A e^(2x) + 3B e^(3x).

  • y(0) = 1 gives A + B = 1.
  • y′(0) = 0 gives 2A + 3B = 0.

Solving: B = −2 and A = 3, so y = 3e^(2x) − 2e^(3x).

Check: y(0) = 3 − 2 = 1, and y′(0) = 6 − 6 = 0.

Non-Homogeneous Equations

The general solution of a y″ + b y′ + c y = f(x) is:

y = complementary function (CF) + particular integral (PI)

  • The complementary function is the general solution of the homogeneous equation (with f(x) replaced by 0), and it contains the two constants.
  • The particular integral is any one solution of the full equation, with no arbitrary constants.

The method of undetermined coefficients

Guess a PI of the same form as f(x), with unknown coefficients, substitute it, and compare coefficients:

  • If f(x) is a polynomial of degree n, try a general polynomial of degree n.
  • If f(x) = k e^(px), try C e^(px).
  • If f(x) = k sin px or k cos px, try C cos px + D sin px.

Worked example 5: a polynomial forcing term

Solve y″ − 3y′ + 2y = 4x.

CF: m² − 3m + 2 = 0 gives m = 1 or 2, so the CF is A eˣ + B e^(2x).

PI: try y = px + q, so y′ = p and y″ = 0. Substituting: 0 − 3p + 2(px + q) = 4x, so 2px + (2q − 3p) = 4x.

Comparing coefficients: 2p = 4, so p = 2, and 2q − 3p = 0, so q = 3. The PI is 2x + 3.

General solution: y = A eˣ + B e^(2x) + 2x + 3.

Worked example 6: an exponential forcing term

Solve y″ + y = e^(2x).

CF: m² + 1 = 0 gives m = ±i, so the CF is A cos x + B sin x.

PI: try y = C e^(2x), so y″ = 4C e^(2x). Then 4C + C = 1, so C = ⅕.

General solution: y = A cos x + B sin x + ⅕ e^(2x).

Worked example 7: a trigonometric forcing term

Solve y″ + 4y = sin x.

CF: m² + 4 = 0 gives m = ±2i, so the CF is A cos 2x + B sin 2x.

PI: try y = C cos x + D sin x, so y″ = −C cos x − D sin x. Then 3C cos x + 3D sin x = sin x, so C = 0 and D = ⅓.

General solution: y = A cos 2x + B sin 2x + ⅓ sin x.

Resonance: when the guess fails

If the trial PI is already part of the CF, substituting it gives 0, which cannot match f(x). Multiply the trial function by x and try again.

For example, for y″ + 4y = sin 2x, the CF already contains sin 2x and cos 2x. Trying x(C cos 2x + D sin 2x) gives the PI −(x/4) cos 2x. Its amplitude grows steadily with x. Physically, this is resonance: forcing a system at its natural frequency makes the oscillations grow without limit. That is why soldiers break step when crossing a bridge.

Application: Simple and Damped Harmonic Motion

A mass m on a spring of stiffness k, displaced by x, obeys Newton's second law:

m x″ + kx = 0, or x″ + ω²x = 0 with ω = √(k/m).

The auxiliary equation m² + ω² = 0 has roots ±ωi, so x = A cos ωt + B sin ωt. This is simple harmonic motion, with period T = 2π/ω.

Worked example 8

A 0.5 kg mass on a spring with k = 50 N/m is pulled 0.1 m from equilibrium and released from rest. Find its motion.

ω = √(50 ÷ 0.5) = 10 rad/s. With x(0) = 0.1 and x′(0) = 0, A = 0.1 and B = 0, so x = 0.1 cos 10t. The period is 2π/10 ≈ 0.63 s.

Damping

Adding friction (a resistance proportional to velocity) gives m x″ + c x′ + kx = 0. The three cases of the auxiliary equation then have physical meanings:

  • Complex roots (light damping): the system oscillates with decreasing amplitude, like a plucked guitar string.
  • Repeated root (critical damping): it returns to equilibrium as fast as possible without oscillating. Car shock absorbers are designed to be close to this.
  • Distinct real roots (heavy damping): it returns slowly without oscillating, like a door closer.

Common Mistakes with Second-Order Differential Equations

  • Writing A e^(mx) + B e^(mx) for a repeated root. These are the same function. The second solution needs the factor x.
  • Forgetting the e^(αx) factor with complex roots, or using the wrong sign for α.
  • Applying the initial conditions to the CF alone in a non-homogeneous equation. Find the full general solution (CF + PI) first.
  • Not multiplying by x when the trial PI is already part of the CF.
  • Trying only sin px when f(x) = sin px. The trial must include both cos px and sin px, unless there is no first-derivative term.

Second-Order Differential Equations: Practice Questions

  1. Solve y″ − y′ − 6y = 0.
  2. Solve y″ + 6y′ + 9y = 0.
  3. Solve y″ + 9y = 0, given that y(0) = 2 and y′(0) = 3.
  4. Solve y″ − 4y′ + 13y = 0.
  5. Find a particular integral of y″ + 3y′ + 2y = 6.
  6. Find a particular integral of y″ − y = 3e^(2x).

Answers: 1. m = 3 or −2, so y = A e^(3x) + B e^(−2x). 2. m = −3 (repeated), so y = (A + Bx)e^(−3x). 3. y = A cos 3x + B sin 3x. Then A = 2, and 3B = 3 gives B = 1, so y = 2 cos 3x + sin 3x. 4. m = 2 ± 3i, so y = e^(2x)(A cos 3x + B sin 3x). 5. Try a constant k: 2k = 6, so the PI is y = 3. 6. Try C e^(2x): 4C − C = 3, so C = 1 and the PI is e^(2x).

Conclusion

Linear second-order differential equations with constant coefficients, a y″ + b y′ + c y = f(x), are solved through the auxiliary equation am² + bm + c = 0. Distinct real roots give A e^(m₁x) + B e^(m₂x), a repeated root gives (A + Bx)e^(mx), and complex roots α ± βi give e^(αx)(A cos βx + B sin βx). For a non-homogeneous equation, add a particular integral found by undetermined coefficients, multiplying by x in the resonance case. Two initial conditions fix the constants. These equations describe springs, pendulums, circuits and damping throughout physics and engineering.

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