MathsDifferential equations › Second order equations

Second order equations

Guess e to the mx, and a differential equation collapses to a quadratic: its two roots, real, repeated or complex, dictate everything the solution can do.

Year FMEDEXCEL 9FM0 CP2

Builds on First order equations and integrating factors and Complex arithmetic and the Argand diagram.

IN THIS TOPIC

  • Reduce ay'' + by' + cy = 0 to its auxiliary quadratic and classify the roots.
  • Write the general solution in each of the three root cases.
  • Use two initial conditions to pin both constants.

WHAT YOU PROBABLY THINK

A differential equation with a second derivative in it is a whole order of magnitude harder than a quadratic.

The auxiliary quadratic

Try y = e^(mx) in ay'' + by' + cy = 0: every term carries e^(mx), which cancels, leaving the auxiliary equation:

am2 + bm + c = 0

The differential equation is exactly as hard as this quadratic, no more, and the opening claim collapses with it. Distinct real roots m₁ and m₂ give y = Ae^(m₁x) + Be^(m₂x); a repeated root m gives y = (A + Bx)e^(mx); complex roots p ± qi give y = e^(px)(A cos qx + B sin qx), an oscillation inside an exponential.

Three root cases, three behaviours: real decay, the repeated borderline, and complex-root oscillationtwo real rootsrepeated rootcomplex rootsy'' , y' and y in balance: the auxiliary roots decide
FIG. 1The three faces of ay'' + by' + cy = 0: two real decays, the repeated borderline with its extra factor of x, and a complex-root oscillation.

WORKED EXAMPLE

Distinct roots with initial conditions

Solve y'' − 5y' + 6y = 0 with y(0) = 0 and y'(0) = 1.

Auxiliary: m² − 5m + 6 = (m − 2)(m − 3) = 0, so m = 2, 3.

General solution: y = Ae2x + Be3x.

y(0) = 0: A + B = 0. y'(0) = 1: 2A + 3B = 1. So B = 1, A = −1.

y = e3x − e2x: two conditions, two constants, both used.

When the roots turn complex

Complex roots do not signal a mistake; they signal oscillation. For y'' + 4y = 0 the auxiliary m² + 4 = 0 gives m = ±2i, so y = A cos 2x + B sin 2x: pure oscillation at angular frequency 2, the differential equation behind simple harmonic motion. The real part of the root controls growth or decay; the imaginary part sets the frequency.

y = cos 2x solves y'' + 4y = 0: acceleration equals minus four times displacement, so it swings for everπamplitude 1y = cos 2xperiod π: frequency 2 from m = ±2i
FIG. 2y = cos 2x solves y'' + 4y = 0: acceleration proportional to minus displacement, the signature of simple harmonic motion.

YOUR TURN

A repeated root

Solve y'' − 4y' + 4y = 0.

Show the working

Auxiliary: m² − 4m + 4 = (m − 2)² = 0, repeated root m = 2.

The repeated case needs its extra x: y = (A + Bx)e2x.

Without the Bx term the 'two' solutions would be one in a wig, and two initial conditions could not both be met.

THE EXAM BIT

  • Write the auxiliary equation down as a labelled step; the method mark attaches to it.
  • The repeated-root case must carry (A + Bx); forgetting the x is the classic error.
  • For complex roots p ± qi, e^(px) takes the real part and cos/sin take the imaginary part.
  • Differentiate the general solution before substituting a y' condition, not after.

CHECK YOURSELF

Find the roots of the auxiliary equation for y'' + y' − 6y = 0 and write the general solution.

Show a hint

m² + m − 6 factorises.

Show the answer

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Substitute e^(mx): the equation becomes am² + bm + c = 0, and the roots classify the motion.

Real roots decay or grow; repeated roots need (A + Bx); complex p ± qi oscillate at frequency q inside e^(px).

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Reduce ay'' + by' + cy = 0 to its auxiliary quadratic and classify the roots.
  • Write the general solution in each of the three root cases.
  • Use two initial conditions to pin both constants.

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