Constant coefficients

The previous lesson proved that a second order linear equation has exactly two independent solutions and gave no way to find even one, and this lesson closes that gap for the case that covers most of physics.

The case is constant coefficients:

a y'' + b y' + c y = 0

with $a$ , $b$ and $c$ numbers rather than functions of $t$ . Every mass on a spring, every RLC circuit, every small oscillation about equilibrium in a system whose parts do not change with time is of this form, and the whole problem reduces to solving a quadratic.

The guess

Look at what the equation asks. A solution must be a function that reproduces itself, up to a constant, when differentiated once and again, so that the three terms can cancel. Exponentials do exactly that, so try $y = e^{r t}$ with $r$ unknown.

Then $y' = r e^{r t}$ and $y'' = r^{2} e^{r t}$ , and substituting gives

(a r^{2} + b r + c) e^{r t} = 0

The exponential is never zero, so the trial function solves the equation precisely when

a r^{2} + b r + c = 0

the characteristic equation. A calculus problem has become an algebra problem, and this substitution is the entire method. The quadratic has two roots, and the three cases of the discriminant $b^{2} - 4 a c$ give the three behaviours that the rest of the course keeps meeting.

Two real roots

If $b^{2} - 4 a c > 0$ the roots $r_{1} \neq r_{2}$ are real, and $e^{r_{1} t}$ and $e^{r_{2} t}$ are two solutions. They are independent: their Wronskian is $(r_{2} - r_{1}) e^{(r_{1} + r_{2}) t}$ , which is nonzero exactly because the roots differ. So the general solution is

y = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t}

and no other solutions exist, by the dimension argument of the previous lesson.

Example. Solve $y'' + 5 y' + 6 y = 0$ with $y (0) = 2$ and $y' (0) = - 1$ .

The characteristic equation is $r^{2} + 5 r + 6 = (r + 2) (r + 3) = 0$ , so $r = - 2$ and $r = - 3$ and $y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t}$ . The conditions give $c_{1} + c_{2} = 2$ and $- 2 c_{1} - 3 c_{2} = - 1$ . Substituting $c_{2} = 2 - c_{1}$ into the second gives $c_{1} - 6 = - 1$ , so $c_{1} = 5$ and $c_{2} = - 3$ , and

y = 5 e^{- 2 t} - 3 e^{- 3 t}

At $t = 1$ this is $5 (0.1353) - 3 (0.0498) = 0.527$ . Both terms decay, so $y \to 0$ , and after a short time the $e^{- 3 t}$ term is negligible and the decay is governed by the root nearer zero. That root is the one that matters in any real system: the slowest decaying mode dominates everything eventually.

Now you. Solve $y'' + y' - 12 y = 0$ with $y (0) = 1$ and $y' (0) = 10$ .

Answer

$r^{2} + r - 12 = (r - 3) (r + 4) = 0$ , so $y = c_{1} e^{3 t} + c_{2} e^{- 4 t}$ . Then $c_{1} + c_{2} = 1$ and $3 c_{1} - 4 c_{2} = 10$ , giving $7 c_{1} = 14$ , so $c_{1} = 2$ and $c_{2} = - 1$ , and $y = 2 e^{3 t} - e^{- 4 t}$ . The positive root means this solution grows without bound.

Complex roots and Euler's formula

If $b^{2} - 4 a c < 0$ the roots are a complex conjugate pair $r = α \pm i β$ , with $α = - b / 2 a$ and $β = \sqrt{4 a c - b^{2}} / 2 a$ . The algebra does not care: $e^{(α + i β) t}$ is a perfectly good solution once the exponential of a complex number is defined. But the equation has real coefficients and describes a real system, so a real answer is wanted.

Euler's formula, $e^{i θ} = \cos θ + i \sin θ$ , supplies it. Writing

e^{(α + i β) t} = e^{α t} (\cos β t + i \sin β t)

and using superposition, the sum of the two complex solutions is $2 e^{α t} \cos β t$ and their difference is $2 i e^{α t} \sin β t$ . Both are combinations of solutions, so both are solutions, and dividing by the constants leaves two real ones. Their Wronskian is $β e^{2 α t}$ , nonzero since $β \neq 0$ , so the general solution is

y = e^{α t} (c_{1} \cos β t + c_{2} \sin β t)

This is the most important formula in the subject. The real part of the root sets the envelope, growing if $α > 0$ and decaying if $α < 0$ , and the imaginary part sets the frequency of oscillation inside it. A single complex number carries both facts, which is why engineers work with complex roots and read off the physics at the end.

Example. Solve $y'' + 2 y' + 5 y = 0$ with $y (0) = 1$ and $y' (0) = 3$ , and give the amplitude of the oscillation at $t = 0$ .

The characteristic equation is $r^{2} + 2 r + 5 = 0$ , with roots $r = \frac{- 2 \pm \sqrt{4 - 20}}{2} = - 1 \pm 2 i$ . So $y = e^{- t} (c_{1} \cos 2 t + c_{2} \sin 2 t)$ . At $t = 0$ , $y = c_{1} = 1$ . Differentiating with the product rule,

y' = - e^{- t} (c_{1} \cos 2 t + c_{2} \sin 2 t) + e^{- t} (- 2 c_{1} \sin 2 t + 2 c_{2} \cos 2 t)

so $y' (0) = - c_{1} + 2 c_{2} = 3$ , giving $c_{2} = 2$ . The solution is $y = e^{- t} (\cos 2 t + 2 \sin 2 t)$ . Combining the trigonometric pair, its initial amplitude is $\sqrt{1 + 4} = 2.236$ , so the motion is an oscillation of period $π$ inside an envelope $\pm 2.236 e^{- t}$ . At $t = 1$ the value is $0.516$ .

Now you. Solve $y'' + 4 y' + 13 y = 0$ with $y (0) = 2$ and $y' (0) = - 1$ .

Answer

The roots are $r = \frac{- 4 \pm \sqrt{16 - 52}}{2} = - 2 \pm 3 i$ , so $y = e^{- 2 t} (c_{1} \cos 3 t + c_{2} \sin 3 t)$ . Then $c_{1} = 2$ , and $y' (0) = - 2 c_{1} + 3 c_{2} = - 1$ gives $c_{2} = 1$ . So $y = e^{- 2 t} (2 \cos 3 t + \sin 3 t)$ .

Writing the answer as a single sinusoid is often clearer. Since $c_{1} \cos β t + c_{2} \sin β t = R \cos (β t - δ)$ with $R = \sqrt{c_{1}^{2} + c_{2}^{2}}$ and $\tan δ = c_{2} / c_{1}$ , the example above is $y = 2.236 e^{- t} \cos (2 t - 1.107)$ , with the phase in radians. The amplitude-and-phase form is what an oscilloscope shows; the sine-and-cosine form is what the initial conditions fit easily. Both are used, and converting between them is worth doing until it is automatic.

The repeated root

If $b^{2} - 4 a c = 0$ there is only one root, $r = - b / 2 a$ , and only one solution $e^{r t}$ . The theory demands two, so the second must be found another way, and reduction of order supplies it exactly as in the previous lesson: putting $y = v e^{r t}$ into the equation makes the terms in $v$ and $v'$ both vanish, leaving $v'' = 0$ , so $v = A + B t$ and the new solution is $t e^{r t}$ . The general solution is

y = (c_{1} + c_{2} t) e^{r t}

The factor of $t$ is not a patch. It is what the double root means: the two exponentials of the nearby non-repeated case, $e^{r_{1} t}$ and $e^{r_{2} t}$ , have a difference which, divided by $r_{2} - r_{1}$ and taken to the limit as the roots merge, is precisely $t e^{r t}$ . The pair does not degenerate into one solution; it degenerates into a solution and a derivative.

Example. Solve $y'' + 6 y' + 9 y = 0$ with $y (0) = 1$ and $y' (0) = - 1$ .

The characteristic equation is $r^{2} + 6 r + 9 = (r + 3)^{2}$ , so $r = - 3$ twice and $y = (c_{1} + c_{2} t) e^{- 3 t}$ . Then $c_{1} = 1$ , and differentiating, $y' = c_{2} e^{- 3 t} - 3 (c_{1} + c_{2} t) e^{- 3 t}$ , so $y' (0) = c_{2} - 3 c_{1} = - 1$ and $c_{2} = 2$ . Hence $y = (1 + 2 t) e^{- 3 t}$ , which rises briefly before decaying, since the linear factor beats the exponential at first. At $t = 1$ it is $3 e^{- 3} = 0.149$ .

Now you. Solve $y'' - 4 y' + 4 y = 0$ with $y (0) = 3$ and $y' (0) = 4$ .

Answer

$r^{2} - 4 r + 4 = (r - 2)^{2}$ , so $y = (c_{1} + c_{2} t) e^{2 t}$ . Then $c_{1} = 3$ and $y' (0) = c_{2} + 2 c_{1} = 4$ , so $c_{2} = - 2$ and $y = (3 - 2 t) e^{2 t}$ .

Reading the roots

Before any initial conditions are applied, the roots already say what the system does, and this is the habit worth acquiring.

Both roots negative and real: the solution decays without oscillating, and the slower root sets the timescale. Roots complex with negative real part: a decaying oscillation, frequency $β$ , envelope time constant $1 / | α |$ . Purely imaginary roots, meaning $b = 0$ : undamped oscillation at frequency $β$ forever. Any root with a positive real part: growth, and the system is unstable, oscillating or not. A zero root, meaning $c = 0$ : a constant solution, so the system has no restoring force and drifts.

That classification is the whole qualitative theory of a linear second order system, and the next lesson does nothing but attach physical names to its cases.

Higher order, and one other solvable family

The method does not care about the order. For $a_{n} y^{(n)} + \dots + a_{0} y = 0$ , substituting $e^{r t}$ gives a polynomial of degree $n$ , and each root contributes a solution: real roots give exponentials, complex conjugate pairs give $e^{α t} \cos β t$ and $e^{α t} \sin β t$ , and a root repeated $k$ times contributes its exponential multiplied by $1, t, \dots, t^{k - 1}$ . Together they give $n$ independent solutions, and the solution space of an $n$ -th order linear equation has dimension $n$ .

For example $y''' - y'' - 4 y' + 4 y = 0$ has characteristic polynomial $r^{3} - r^{2} - 4 r + 4 = (r - 1) (r - 2) (r + 2)$ , so the general solution is $c_{1} e^{t} + c_{2} e^{2 t} + c_{3} e^{- 2 t}$ . The honest limit is that polynomials of degree five and above have no formula for their roots, so beyond the quartic the characteristic equation itself has to be solved numerically. That is not a serious obstacle, since a numerical root is as good as an exact one for computing a response, but it does mean the method stops being a closed form procedure.

One other family yields to a substitution of the same spirit. The Cauchy-Euler equation $t^{2} y'' + α t y' + β y = 0$ has coefficients that are not constant, but each derivative comes multiplied by exactly the matching power of $t$ , so trying $y = t^{r}$ works: it gives $r (r - 1) + α r + β = 0$ . The equation $t^{2} y'' - 3 t y' + 4 y = 0$ from the previous lesson gives $r^{2} - 4 r + 4 = 0$ , a repeated root at $r = 2$ , and the second solution acquires a factor of $\ln t$ rather than $t$ , which is what reduction of order produced there.

What still cannot be done

Everything here is homogeneous: no forcing, no external driving. All these solutions decay, grow or oscillate on their own terms and then, if stable, sit at zero. A system that is being pushed needs a particular solution, which is a different problem.

It is also worth being clear about the reach of constant coefficients. They describe a system whose properties do not change with time and whose response is proportional to the disturbance. A spring stretched too far, a circuit driven into saturation, a pendulum swung through a large angle: none of these is covered, and the last of them is the subject of the final lesson.

Next, though, comes the physics. The equation $m x'' + c x' + k x = 0$ is one equation with three regimes, and the same three roots that came out of the algebra above turn out to be the difference between a car that rides well and one that bounces down the road.