Second-order equations and oscillation

Chapter 7 - Position needs velocity

A second-order equation tracks curvature or acceleration. To select one motion, you usually need two initial facts, such as position and velocity:

$y''+ay'+by=0,\qquad y(0)=y_0,\quad y'(0)=v_0.$

The solution is built from natural modes: motions the unforced system can sustain on its own.

From derivatives to a polynomial

Exponentials reproduce themselves under differentiation. Try $y=e^{rt}$ :

$r^2e^{rt}+ar e^{rt}+be^{rt}=0.$

Since $e^{rt}\ne0$ , the possible rates $r$ solve the characteristic equation

$r^2+ar+b=0.$

The roots predict the motion's geometry.

Two real roots $r_1\ne r_2$ y=C_1e^{r_1t}+C_2e^{r_2t}

The mode with the larger root dominates late if its coefficient is nonzero. If both roots are negative, this is the slower-decaying mode.

Repeated real root $r$ y=(C_1+C_2t)e^{rt}

The factor $t$ creates a second independent motion.

Complex roots $\alpha\pm i\beta$ y=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t)

$\alpha$ controls the envelope, the curve bounding the oscillation's size; $\beta$ controls angular frequency, measured in radians per unit time.

Root signsRe(r) < 0 → decay

Positive real part gives a growing mode. A nonzero pure imaginary pair gives undamped oscillation; a zero root gives a constant mode, and a repeated zero root also permits linear drift.

Lab: watch damping reshape motion

Damped oscillator

x''+cx'+x=0

Damping c 0.35 Initial position 1.5 Initial velocity 0.0

position over timeposition-velocity path

Move $c$ slowly through 2. For $0\lt c\lt 2$ , roots are complex and nonzero trajectories spiral inward; at $c=0$ , they form closed orbits without decay. At 2, the root repeats and the system returns without sustained oscillation. Above 2, two negative real modes give a non-oscillatory return; the slower mode decays more slowly than at critical damping unless its coefficient is zero. Critical damping is the boundary between oscillatory and non-oscillatory return.

Worked example: read roots, then constants

Solve $y''+4y'+13y=0$ , $y(0)=2$ , $y'(0)=-1$ fully worked

Characteristic equation. $r^2+4r+13=0$ .
Roots. $r=(-4\pm\sqrt{16-52})/2=-2\pm3i$ .
Predict. The motion oscillates at angular frequency 3 under an envelope $e^{-2t}$ . It decays rapidly.
Write the real family. $y=e^{-2t}(C_1\cos3t+C_2\sin3t)$ .
Use position. $y(0)=C_1=2$ .
Use velocity. Differentiating and evaluating at zero gives $y'(0)=-2C_1+3C_2=-1$ . With $C_1=2$ , $C_2=1$ .
Result and check. $y=e^{-2t}(2\cos3t+\sin3t)$ . The roots guarantee the equation; direct evaluation checks both initial conditions.

Fade the root cases

Match each equation to a root picture and solution form.

$y''+5y'+6y=0$

Roots: $\underline{\hspace{5em}}$

Form: $\underline{\hspace{8em}}$

$y''+4y'+4y=0$

Root: $\underline{\hspace{5em}}$

Form: $\underline{\hspace{8em}}$

$y''+9y=0$

Roots: $\underline{\hspace{5em}}$

Interpret the motion.

Check the forms

Roots $-2,-3$ ; $y=C_1e^{-2t}+C_2e^{-3t}$ .
Repeated root $-2$ ; $y=(C_1+C_2t)e^{-2t}$ .
Roots $\pm3i$ ; $y=C_1\cos3t+C_2\sin3t$ , an undamped oscillation.

Mechanical language

The mass-spring-damper model

$mx''+cx'+kx=0$

balances inertia $mx''$ , damping $cx'$ , and restoring force $kx$ . Dividing by $m$ shows that the key comparisons use $c/m$ and $k/m$ . The discriminant $c^2-4mk$ separates overdamped, critically damped, and underdamped motion.

Root reading. A characteristic root is $-0.2+5i$ . What does each part control?

Error clinic

Error clinic: one constant for a second-order equation

A second-order linear homogeneous equation generally needs two independent modes and two constants. A repeated root does not reduce that dimension; it changes the second mode to $te^{rt}$ .

Error clinic: using complex exponentials but dropping real solutions

Complex roots are a calculation device. For real initial data, report the equivalent real sine-cosine form unless the context explicitly prefers complex notation.

Practice

Practice - distinct roots
7.1 Solve $y''-y'-6y=0$ , $y(0)=1$ , $y'(0)=0$ . Predict which mode dominates for large positive time.
Detailed solution
Practice - repeated root
7.2 Solve $y''+6y'+9y=0$ , $y(0)=0$ , $y'(0)=2$ .
Detailed solution
Integrated - oscillator
7.3 For $x''+2\zeta x'+x=0$ , classify motion for $\zeta=0, 1/2, 1, 2$ . Connect each case to a path in position-velocity space.
Detailed solution
Challenge - design
7.4 Choose $a$ and $b$ so $y''+ay'+by=0$ has oscillations with envelope $e^{-t}$ and period $\pi$ . Explain from the desired roots.
Detailed solution
Practice - extension
7.5 Solve $x''+4x=0$ , $x(0)=1$ , $x'(0)=2$ . Rewrite the solution as a single shifted cosine, identify its amplitude and period, and verify that $E=(x')^2/2+2x^2$ stays constant.
Detailed solution
Challenge - transfer
7.6 Solve $y''+2y'=0$ , $y(0)=1$ , $y'(0)=-4$ . Identify both characteristic modes and the limit as $t\to\infty$ . Explain why the zero root produces neither a decaying mode nor a sustained oscillation.
Detailed solution

Retrieve across chapters

Explain how these are the same idea in different dimensions:

a stable equilibrium on a phase line;
a negative real characteristic root; and
an inward-moving position-velocity path.

I predicted motion from roots before solving for constants and connected the two plots in the lab.

Next: Add an external input and learn why matching frequencies can amplify motion.

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Differential Equations as a Second Language