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:

$y^{''} + a y^{'} + b y = 0, y (0) = y_{0}, y^{'} (0) = v_{0} .$

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}):

$r^{2} e^{r t} + a r e^{r t} + b e^{r t} = 0.$

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

$r^{2} + a r + b = 0.$

The roots predict the motion's geometry.

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

Two exponential modes; the slower-decaying mode dominates late.

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; $\beta$ controls angular frequency.

Root signsRe(r) < 0 → decay

Positive real part grows; zero real part sustains an undamped oscillation.

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. Below 2, roots are complex and the path spirals. At 2, the root repeats and the system returns without oscillating. Above 2, two negative real modes return even more slowly. 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

$m x^{''} + c x^{'} + k x = 0$

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

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