6.8 Higher-Order Linear Differential Equations & Engineering Applications

Key Takeaways

  • The general solution of a linear differential equation with constant coefficients is the complementary function plus the particular integral, corresponding physically to the transient and steady-state responses.
  • Distinct real roots of the auxiliary equation give exponential terms, repeated roots introduce an extra factor of x, and complex roots produce damped sinusoidal terms.
  • The particular integral for a right-hand side of e^(ax) is found by substituting a for D, and the case of failure when D minus a vanishes is repaired by multiplying by x.
  • The mass-spring-damper equation m x'' + c x' + k x = F(t) is the physical archetype of this topic and reappears directly in the Vibrations part of Paper-II.
Last updated: August 2026

Structure of the General Solution

A linear differential equation with constant coefficients has the form

andnydxn+an1dn1ydxn1++a0y=X(x)a_n\frac{d^ny}{dx^n} + a_{n-1}\frac{d^{n-1}y}{dx^{n-1}} + \cdots + a_0 y = X(x)

Writing $D \equiv d/dx$ turns this into $f(D)y = X$. The complete solution is always

y=CFcomplementary function+PIparticular integraly = \underbrace{\text{CF}}_{\text{complementary function}} + \underbrace{\text{PI}}_{\text{particular integral}}

The engineering reading of that split is worth fixing in mind now, because it recurs throughout vibration and control work: the complementary function is the transient — the system's own free response, which decays if the system is damped — and the particular integral is the steady state, the part that persists and mirrors the forcing.

The Complementary Function

Set $X = 0$ and substitute $y = e^{mx}$ to obtain the auxiliary equation $f(m) = 0$. The form of the CF depends on the nature of the roots:

Roots of auxiliary equationContribution to CF
Real and distinct $m_1, m_2$$C_1e^{m_1x} + C_2e^{m_2x}$
Real and repeated $m$ (twice)$(C_1 + C_2x)e^{mx}$
Real and repeated $m$ (three times)$(C_1 + C_2x + C_3x^2)e^{mx}$
Complex pair $\alpha \pm i\beta$$e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

The extra powers of $x$ for repeated roots are needed to keep the solutions linearly independent, and forgetting them is the most common source of lost marks.

Example. For $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$, the auxiliary equation $m^2 - 5m + 6 = 0$ gives $m = 2, 3$, so $y = C_1e^{2x} + C_2e^{3x}$.

Example. For $y'' + 4y = 0$, the roots are $m = \pm 2i$, so $\alpha = 0$ and $\beta = 2$, giving $y = C_1\cos 2x + C_2\sin 2x$ — undamped simple harmonic motion of angular frequency 2 rad/s.

Particular Integrals

The PI is $\dfrac{1}{f(D)}X$, evaluated by rules that depend on the form of $X$:

$X(x)$Rule
$e^{ax}$Replace $D$ by $a$; if $f(a) = 0$, multiply by $x$ and differentiate $f$
$\sin ax$ or $\cos ax$Replace $D^2$ by $-a^2$; if that fails, multiply by $x$
$x^m$Expand $\dfrac{1}{f(D)}$ as a binomial series in $D$ up to $D^m$
$e^{ax}V(x)$Shift: $\dfrac{1}{f(D)}e^{ax}V = e^{ax}\dfrac{1}{f(D+a)}V$

The "case of failure" rows matter physically: they are the mathematics of resonance, where the forcing frequency coincides with a natural frequency and the response grows linearly with $x$ rather than staying bounded.

Worked Example: Full Solution

Solve $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = e^{4x}$.

CF. From above, $\text{CF} = C_1e^{2x} + C_2e^{3x}$.

PI. Replace $D$ by 4 in $\dfrac{1}{D^2 - 5D + 6}$:

PI=e4x1620+6=e4x2\text{PI} = \frac{e^{4x}}{16 - 20 + 6} = \frac{e^{4x}}{2}

Complete solution. $y = C_1e^{2x} + C_2e^{3x} + \tfrac{1}{2}e^{4x}$.

Had the forcing been $e^{2x}$ instead, $f(2) = 0$ and the rule fails; the repair is $\text{PI} = x,e^{2x}/f'(2) = x e^{2x}/(2(2)-5) = -xe^{2x}$.

Cauchy-Euler (Homogeneous Linear) Equations

An equation of the form

x2d2ydx2+axdydx+by=Xx^2\frac{d^2y}{dx^2} + a x\frac{dy}{dx} + by = X

has variable coefficients but is reduced to constant coefficients by the substitution $x = e^{z}$, i.e. $z = \ln x$. Then $x\dfrac{dy}{dx} = \dfrac{dy}{dz}$ and $x^2\dfrac{d^2y}{dx^2} = \dfrac{d^2y}{dz^2} - \dfrac{dy}{dz}$, after which the standard method applies. These equations arise in thick-cylinder and rotating-disc stress analysis.

The Physical Archetype: Mass-Spring-Damper

Every mechanical vibration problem in Paper-II is this one equation:

md2xdt2+cdxdt+kx=F(t)m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = F(t)

Its auxiliary equation is $ms^2 + cs + k = 0$, with roots

s=c±c24mk2ms = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}

The discriminant decides the character of the free response, and the classification maps exactly onto the root table above:

DiscriminantRootsPhysical regime
$c^2 > 4mk$Real, distinct, both negativeOver-damped — returns without oscillating
$c^2 = 4mk$Real, repeatedCritically damped — fastest non-oscillatory return
$c^2 < 4mk$Complex pairUnder-damped — decaying oscillation

The critical damping coefficient is therefore $c_c = 2\sqrt{mk} = 2m\omega_n$, and the damping ratio is $\zeta = c/c_c$. The undamped natural frequency is $\omega_n = \sqrt{k/m}$ and the damped natural frequency is $\omega_d = \omega_n\sqrt{1 - \zeta^2}$.

Recognising that the vibration chapter is simply this differential-equation topic wearing engineering clothing saves a great deal of separate memorisation.

Test Your Knowledge

The complementary function of the differential equation y'' + 4y = 0 is:

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Test Your Knowledge

If the auxiliary equation of a second-order linear differential equation has a repeated real root m, the complementary function is:

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D
Test Your Knowledge

For a mass-spring-damper system, the condition c^2 = 4mk corresponds to:

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D
Test Your Knowledge

In the complete solution of a forced linear differential equation, the complementary function physically represents the:

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D