Differential Equations · step 7 of 14

Second-order linear homogeneous equations

For ay″ + by′ + cy = 0 with constant coefficients, substituting y = e^(rx) gives the characteristic equation ar² + br + c = 0. Three cases follow from the discriminant: distinct real roots give Ae^(r₁x) + Be^(r₂x); a repeated root r gives (A + Bx)e^(rx); and complex roots α ± βi give e^(αx)(A cos βx + B sin βx).

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Write the characteristic equationReplace y″ by r², y′ by r and y by 1.
Compute the discriminant and identify the casePositive gives distinct real; zero gives repeated; negative gives complex conjugates.
Write the general solution for that caseRepeated roots need the extra factor of x; complex roots convert to the real trigonometric form.
Interpret physicallyDistinct negative roots mean overdamping; a repeated root means critical damping; complex roots mean oscillation.

Worked example

Solve y″ + 4y′ + 13y = 0.

Characteristic equation: r² + 4r + 13 = 0.
Discriminant: 16 − 52 = −36, so the roots are complex.
r = (−4 ± 6i)/2 = −2 ± 3i, so α = −2 and β = 3.
Apply the complex-root form.

Answer. y = e^(−2x)(A cos 3x + B sin 3x) — a decaying oscillation, the signature of underdamping.

Where marks get dropped

These are the specific errors that cost credit on second-order linear homogeneous equations questions — QED's rubric penalises each of them separately.

Writing Ae^(rx) + Be^(rx) for a repeated root. Those merge into one constant; the second solution is xe^(rx).
Leaving the answer in complex exponential form when a real solution is wanted. Convert using Euler’s formula.
Mixing up α and β. α is the real part (the decay rate) and β the imaginary part (the frequency).

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Second-order linear homogeneous equations — frequently asked questions

Why does a repeated root give an extra factor of x?

Because one exponential provides only one independent solution, and a second-order equation needs two. Reduction of order produces xe^(rx) as the second.

How do complex exponentials become sines and cosines?

By Euler’s formula: e^((α+βi)x) = e^(αx)(cos βx + i sin βx). Taking real and imaginary parts gives two real independent solutions.

What if the coefficients are not constant?

The characteristic method fails. Cauchy–Euler equations (with x² y″ terms) have their own substitution; otherwise series solutions are needed.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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