Second-Order Linear ODEs

Push a child on a swing once and let go. It swings back and forth, but each swing is a little smaller than the last until it finally stops. A weight bouncing on a spring does exactly the same thing. The "back-and-forth" comes from the spring; the "slowly dies down" comes from friction and air resistance. Almost everything on this page is just a careful description of that one familiar picture.

Three things act on the mass. The spring pulls it back toward the middle, harder the further you stretch it: a force of $-ky$, where $k$ is the spring's stiffness and $y$ is how far it is from rest. A damper (a shock absorber, or just air) pushes against whatever direction it is moving: a force of $-by'$, where $y'$ is the velocity. And Newton said force equals mass times acceleration, $my''$. Putting the pushes together gives the equation this whole page studies: $my'' + by' + ky = 0$. With no damper ($b=0$) the mass bounces forever at its natural frequency $\omega_0 = \sqrt{k/m}$. For $m=1$ and $k=4$ that is $\omega_0 = 2$ radians per second, so one full bounce takes $T = 2\pi/\omega_0 \approx 3.14$ seconds.

How fast the wiggles die out is set by the damping ratio $\zeta = \dfrac{b}{2\sqrt{mk}}$. Three different behaviors fall out of it. If $\zeta < 1$ (underdamped) the mass still oscillates, but trapped inside a shrinking envelope $e^{-(b/2m)t}$. If $\zeta = 1$ (critically damped) it slides straight back to rest as fast as possible with no overshoot. If $\zeta > 1$ (overdamped) it crawls back slowly without ever crossing zero. Now push the mass rhythmically with $F\cos(\omega t)$: when your push frequency $\omega$ lands near $\omega_0$, the swings build up enormously -- that is resonance. The sliders map straight onto these symbols: $m$, $k$, and $b$ set the character of the motion, while $F$ and the drive frequency $w$ switch on the forcing.

Set $b = 0$ and watch the curve oscillate forever while the energy graph stays perfectly flat -- nothing is lost. Then drag $b$ up past the critical value ($b = 2\sqrt{mk}$) and watch the oscillation disappear into a smooth return. Finally turn $F$ up and slide the drive frequency $w$ toward $\sqrt{k/m}$: the amplitude swells dramatically as you tune into resonance.

$\zeta=b/(2\sqrt{mk})$. Underdamped: $\zeta<1$ (oscillates). Critical: $\zeta=1$ (fastest non-oscillatory return). Overdamped: $\zeta>1$ (slow monotone return).

With forcing $F\cos(\omega t)$: particular solution amplitude $A=F/\sqrt{(k-m\omega^2)^2+(b\omega)^2}$. At $\omega=\omega_0=\sqrt{k/m}$ with $b=0$: $A\to\infty$ -- resonance. With $b>0$: maximum at $\omega_r=\sqrt{k/m-b^2/(2m^2)}$.

$y''+4y'+4y=0$ (critical, $b^2=4mk$). Char. eq. $(r+2)^2=0$, $r=-2$. General: $y=(C_1+C_2t)e^{-2t}$. With $y(0)=1,y'(0)=0$: $C_1=1,C_2=2$. So $y=(1+2t)e^{-2t}$.

KE $= \frac{1}{2}my'^2$, PE $=\frac{1}{2}ky^2$. Total $E=$ KE+PE. For $b=0$: $E=$ const (conserved). For $b>0$: $dE/dt=-by'^2\leq0$ (dissipation). At resonance with forcing: $E$ grows without bound.

For $y''+2y'+5y=3\cos t$: try $y_p=A\cos t+B\sin t$. Substituting: $4A+2B=3$, $-2A+4B=0$. Solving: $A=3/5$, $B=3/10$. General: $y=e^{-t}(C_1\cos2t+C_2\sin2t)+\frac{3}{5}\cos t+\frac{3}{10}\sin t$.

$W(y_1,y_2)=y_1y_2'-y_1'y_2$. If $W\neq0$ at any point then $y_1,y_2$ are linearly independent -- forming a fundamental set. Abel's theorem: $W(t)=W(t_0)\exp(-\int p(t)\,dt)$ where $p(t)$ is the coefficient of $y'$.