The general form of an nth-order linear differential equation with constant coefficients is:
A
a₀y + a₁y' + a₂y'' + ... + aₙyⁿ = 0
Analysis & Theory
The general form is a₀y + a₁y' + a₂y'' + ... + aₙyⁿ = 0, where a₀, a₁, ... aₙ are constants.
To solve an nth-order linear differential equation with constant coefficients, we first form:
A
Auxiliary (or characteristic) equation
B
Particular solution directly
Analysis & Theory
We form the auxiliary or characteristic equation by replacing yⁿ with mⁿ.
The auxiliary equation is obtained by substituting:
Analysis & Theory
We assume y = e^(mx), so derivatives reduce to powers of m.
If the auxiliary equation has distinct real roots m₁, m₂, ..., mₙ, then the general solution is:
A
y = C₁e^(m₁x) + C₂e^(m₂x) + ... + Cₙe^(mₙx)
B
y = (C₁ + C₂ + ... + Cₙ)e^x
C
y = C₁x^m₁ + ... + Cₙx^mₙ
Analysis & Theory
Each distinct root contributes a term of the form C·e^(mx) to the solution.
If the auxiliary equation has repeated roots, the solution involves:
B
Exponential and polynomial terms
Analysis & Theory
For repeated roots, polynomial factors in x multiply the exponential terms.
If the auxiliary equation has complex roots α ± iβ, the solution contains:
Analysis & Theory
Complex roots give terms like e^(αx)(C₁cosβx + C₂sinβx).
The complete solution of a non-homogeneous equation is:
A
Only the complementary function
B
Only the particular integral
C
Complementary function + Particular integral
Analysis & Theory
Complete solution = CF (solution of homogeneous part) + PI (solution of non-homogeneous part).
The solution of y'' - 5y' + 6y = 0 is:
A
y = C₁e^(2x) + C₂e^(3x)
Analysis & Theory
Auxiliary equation: m² - 5m + 6 = 0 ⇒ (m-2)(m-3)=0 ⇒ roots 2,3 ⇒ y = C₁e^(2x) + C₂e^(3x).
The auxiliary equation of y''' + 3y'' + 3y' + y = 0 is:
Analysis & Theory
Replace y, y', y'', y''' with 1, m, m², m³ respectively ⇒ m³ + 3m² + 3m + 1 = 0.
The solution of y'' + y = 0 is:
Analysis & Theory
Auxiliary equation: m² + 1 = 0 ⇒ m = ±i ⇒ y = C₁cosx + C₂sinx.