Constant coefficients and the characteristic equation

Second-order equations

Lesson 6.1.1 called an equation for an unknown function $y(t)$ and its derivatives a differential equation, and its order the highest derivative in it. Module 6.1 solved first-order equations such as $y' = ky$, whose solutions are $e^{kt}$. This lesson solves equations with a second derivative $y''$.

Lesson 2.3.5 met one already, written $f'' = -f$: the second derivative of $\sin t$ is $-\sin t$. The equation says the curve always bends back towards 0, by as much as it is away from 0. That makes it swing.

This lesson solves the family

where $a$, $b$ and $c$ are numbers and $a \neq 0$. It is linear: $y$, $y'$ and $y''$ each appear on their own, to the first power, times a constant. It has constant coefficients: $a$, $b$ and $c$ do not change with $t$, and the right-hand side is 0.

So $y'' + 5y' + 4y = 0$ belongs to the family. $yy'' = 1$ (a product), $(y')^2 = y$ (a power) and $ty'' + y = 0$ (a coefficient that changes) do not.

Two starting values

A first-order equation needs one starting value, $y(0)$. A second-order equation fixes how the curve bends, so a solution also needs its starting slope $y'(0)$: where it starts, and which way it heads. Lesson 6.2.2 uses those two values to pick one solution.

Units. If $t$ is in years, $y'$ is in units of $y$ per year and $y''$ per year squared. With $a = 1$, $b$ is then per year and $c$ per year squared.

Checking a solution. As in lesson 6.1.1, substitute it. For $y = e^{-t}$, $y' = -e^{-t}$ and $y'' = e^{-t}$, so $y'' + 5y' + 4y = (1 - 5 + 4)e^{-t} = 0$. It is a solution.

Trying an exponential

The derivatives of $e^{rt}$ are multiples of itself: $y' = re^{rt}$ and $y'' = r^2e^{rt}$ (lesson 2.3.4). Put $y = e^{rt}$ into $ay'' + by' + cy = 0$ and every term carries the same factor:

Since $e^{rt}$ is never 0, the bracket must be. That quadratic in $r$, $ar^2 + br + c = 0$, is the characteristic equation of the differential equation, and it decides everything.

For $y'' + 5y' + 4y = 0$ the characteristic equation is $r^2 + 5r + 4 = 0$. A table of trial rates finds the two that work:

The quadratic formula of lesson 1.2.2 gives the same pair: $r = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2}$, so $r = -1$ or $r = -4$. Both $e^{-t}$ and $e^{-4t}$ solve the equation; $e^{-2t}$ does not.

Each term of the equation is a constant times $y$ or one of its derivatives. So a sum such as $2e^{-t} - 3e^{-4t}$ is a solution too. Lesson 6.2.2 shows that every solution is such a combination, and how to choose it.

The discriminant $b^2 - 4ac$ of lesson 1.2.2 sorts the equations into three cases: two real roots, one repeated root, or two complex roots.

Two real roots

When $b^2 - 4ac > 0$ there are two real roots $r_1$ and $r_2$, and two solutions $e^{r_1t}$ and $e^{r_2t}$. The signs of the roots say what they do.

• Both negative: both solutions die away. In $y'' + 5y' + 4y = 0$, $e^{-4t}$ fades fastest. At $t = 2$, $e^{-t} = 0.1353$ but $e^{-4t} = 0.000335$, so the root nearer 0 sets the long-run pace.
• One positive: $y'' - y = 0$ has $r^2 - 1 = 0$, so $r = \pm 1$. The solution $e^{t}$ grows without bound and $e^{-t}$ dies away.
• A zero root: $y'' + 2y' = 0$ has $r^2 + 2r = r(r + 2) = 0$, so $r = 0$ or $-2$. The root 0 gives $e^{0t} = 1$: the constant 1 is a solution.

The rates are per unit of $t$: with $t$ in years, $e^{-4t}$ shrinks by the factor $e^{-4}$ every year.

A repeated root

When $b^2 - 4ac = 0$, the two roots coincide. For $y'' + 4y' + 4y = 0$, $r^2 + 4r + 4 = (r + 2)^2$, so $r = -2$ only. That gives one solution, $e^{-2t}$, but a second-order equation needs two.

The second is $te^{-2t}$. Check it with the product rule (lesson 2.3.2):

Then $y'' + 4y' + 4y = \big[(4t - 4) + 4(1 - 2t) + 4t\big]e^{-2t} = 0$. In numbers:

Complex roots: waves

Start with $y'' = -y$, or $y'' + y = 0$. Its characteristic equation $r^2 + 1 = 0$ has no real roots, but it has two complex ones, $r = \pm i$ (lesson 1.7.2). By Euler's formula (lessons 1.7.3 and 2.7.2),

Its real part, $\cos t$, and its imaginary part, $\sin t$, are each solutions. Lesson 2.3.5 found this directly: differentiating $\sin t$ twice gives $-\sin t$, and $\cos t$ gives $-\cos t$. The solutions are waves of period $2\pi$.

The same holds for $y'' + 4y = 0$. Now $r = \pm 2i$, and $e^{2it} = \cos 2t + i\sin 2t$, so the solutions are $\cos 2t$ and $\sin 2t$. They swing twice as fast, with period $\frac{2\pi}{2} = \pi$.

Damped waves

Now add a first-derivative term: $y'' + 2y' + 5y = 0$. The discriminant is $4 - 20 = -16$, and the quadratic formula gives

Split $e^{(-1 + 2i)t}$ by the exponent rules (lesson 1.4.1), then use Euler's formula:

Because $a$, $b$ and $c$ are real, the real part and the imaginary part of a complex solution are each real solutions. So the equation has $e^{-t}\cos 2t$ and $e^{-t}\sin 2t$. The other root, $-1 - 2i$, gives the same pair again.

The waves of $e^{-t}\cos 2t$ swing between the curves $e^{-t}$ and $-e^{-t}$, called the envelope. With $t$ in years, $\omega = 2$ is in radians per year (lesson 2.3.5), and one swing takes $\pi \approx 3.14$ years.

Damping

The $by'$ term is called damping: with $a$ and $c$ positive, a positive $b$ drains the swing. Hold $c = 4$ and change $b$ in $y'' + by' + 4y = 0$:

Small damping lets the waves die down slowly. At $b = 4$ the waves just disappear.

More damping than that gives two negative real roots, and the one nearer 0 makes the return slower. A negative $b$ feeds the swing instead, and the waves grow.

The explorer solves $y'' + by' + cy = 0$ from $y(0) = 1$ and $y'(0) = 0$; its other presets are rows of the damping table. Pick the preset $y'' = -y$ and raise $b$ from 0: the waves die down faster, and from $b = 2$ on they no longer cross zero. Then drag $b$ below 0, and make $c$ negative.

With $a = 1$, the point $(b, c)$ tells you the case at a glance. The roots are complex above the parabola $c = \frac{b^2}{4}$, where $b^2 - 4c < 0$, and real below it. Below the $b$-axis, $c < 0$ makes $\sqrt{b^2 - 4c}$ larger than $|b|$, so one root is positive.

Checking with solve_ivp

Lesson 6.1.2 solved first-order equations numerically with solve_ivp. A second-order equation becomes two first-order ones if the slope gets its own name, $v = y'$:

The second line is the equation $ay'' + by' + cy = 0$ solved for $y''$. Now the unknown is the pair $(y, v)$, which solve_ivp handles as a list.

Lesson 6.2.2 combines the two basic solutions to meet given starting values, or conditions at two points. Case study 6.2.4 uses the same trial-solution idea to value a put that never expires.

Common mistakes

1. Put each coefficient in its place. $y'' - 9y = 0$ has no $y'$ term, so its characteristic equation is $r^2 - 9 = 0$, with roots $\pm 3$. It is not $r^2 - 9r = 0$.
2. Keep $a$. $2y'' + 3y' + y = 0$ gives $2r^2 + 3r + 1 = 0$, with roots $-\frac{1}{2}$ and $-1$.
3. A repeated root gives one exponential. The second solution is $te^{r_0t}$, not $e^{r_0t}$ again. np.roots can report a repeated root as a pair with a tiny imaginary part, such as $10^{-8}$; decide the case from $b^2 - 4ac$.
4. Complex roots: $\alpha$ for size, $\omega$ for speed. For $-1 \pm 2i$ the solutions are $e^{-t}\cos 2t$ and $e^{-t}\sin 2t$, not $e^{2t}\cos t$. The waves die down only if $\alpha < 0$.
5. The period is $\frac{2\pi}{\omega}$, in radians. $y'' + 4y = 0$ has $\omega = 2$ and period $\pi$.
6. The trial $e^{rt}$ needs constant coefficients. In $ty'' + y = 0$ the factor $t$ stays, and no single $r$ works. Lesson 6.2.2 tries powers $x^p$ for a different family.
7. With $a = 0$ there is no second derivative. The equation is first order, as in module 6.1.