Second-Order Differential Equations

Revise Second-Order Differential Equations for Further Pure Mathematics 2 WFM02 (A2 Level) — revision notes and instant AI marking.

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Edexcel IAL Further Pure Mathematics 2 (WFM02)

Second-Order Differential Equations

The big idea: The general solution is always complementary function + particular integral: the CF solves the equation with zero on the right, and the PI is any one solution of the full equation.

Summary — What This Chapter Covers

Auxiliary equation am² + bm + c = 0 decides the complementary function.
Distinct real roots α, β: y = Ae^αx + Be^βx. Equal roots α: y = (A + Bx)e^αx. Complex roots p ± qi: y = e^px(A cos qx + B sin qx).
The particular integral copies the form of f(x); multiply by x if that form already appears in the CF.
General solution = CF + PI; use two boundary conditions to find A and B.
A given substitution (such as x = e^u) reduces some equations to constant-coefficient form.

1. The Complementary Function

Complementary function

Solving ay″ + by′ + cy = 0

Try y = e^mx: it works when am² + bm + c = 0. The three cases of the auxiliary equation give the three forms of the CF.

Roots of the auxiliary equation	Complementary function
Real, distinct α, β	Ae^αx + Be^βx
Real, equal α	(A + Bx)e^αx
Complex p ± qi	e^px(A cos qx + B sin qx)
Pure imaginary ± qi	A cos qx + B sin qx

2. The Particular Integral

Particular integral

Choosing the trial function

f(x)	Trial PI
ke^px	λe^px (λxe^px if p is a root; λx²e^px if a repeated root)
A + Bx or p + qx + cx²	polynomial of the same degree, every term included
m cos ωx + n sin ωx	λ cos ωx + μ sin ωx (x(λ cos ωx + μ sin ωx) if ±iω are roots)

Worked Example

Find the general solution of d²y/dx² − 3 dy/dx + 2y = 4x.

Worked Example

Find the general solution of d²y/dx² + 4y = sin 2x.

3. Boundary Conditions

Particular solution

Finding A and B

Always find the general solution first (CF + PI), then apply the conditions. Two conditions are needed — usually y and dy/dx at one point, or y at two points.

Worked Example

Solve d²y/dx² − 3 dy/dx + 2y = 4x given y = 0 and dy/dx = 0 when x = 0.

4. Substitutions for Second-Order Equations

Substitution

Reducing to constant coefficients

For equations like x² d²y/dx² + ax dy/dx + by = f(x), the substitution x = e^u gives constant coefficients. Use the chain rule carefully:

With x = e^u x dy/dx = dy/du x² d²y/dx² = d²y/du² − dy/du

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Second-Order Differential Equations