6.8 Higher-Order Linear Differential Equations & Engineering Applications

Structure of the General Solution

A linear differential equation with constant coefficients has the form

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

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

$$y = \underbrace{\text{CF}}_{\text{complementary function}} + \underbrace{\text{PI}}_{\text{particular integral}}$$

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:

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 $X$:

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 $\dfrac{1}{D^2 - 5D + 6}$:

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

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

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

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:

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

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

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

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

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.