Second order equations with constant coefficients – Differential Equations (Senior 6 Mathematics) | Ganzaa

Differential Equations

Lessons

Senior 6 · Mathematics

Differential Equations

5 lessons

Lesson 3 of 5

Second order equations with constant coefficients

35 min

Springs, shock absorbers and the second derivative

Press down on the corner of a car and let go. The spring pushes the body up, the shock absorber resists the motion, and the car settles back. Newton's second law (force = mass × acceleration) turns this into an equation with a second derivative: $m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$ , where $x$ is the displacement, $m$ the mass, $c$ the damping constant and $k$ the spring constant. This lesson solves every equation of this shape.

Definition

Second order linear equation with constant coefficients

a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x), \qquad a, b, c \text{ real constants}, \ a \ne 0 .

It is homogeneous if $f (x) = 0$ for all $x$ , and non-homogeneous otherwise (lesson 4).

Characteristic (auxiliary) equation of $a y^{''} + b y^{'} + c y = 0$ : the quadratic $a r^{2} + b r + c = 0$ .

Example: the characteristic equation of $y^{''} - y^{'} - 6 y = 0$ is $r^{2} - r - 6 = 0$ , with roots 3 and $- 2$ .

Where the characteristic equation comes from

Superposition. If $y_{1}$ and $y_{2}$ solve $a y^{''} + b y^{'} + c y = 0$ , so does $C_{1} y_{1} + C_{2} y_{2}$ , because differentiation is linear: the left side becomes $C_1 \times 0 + C_2 \times 0 = 0$ . If $y_{1}$ and $y_{2}$ are not constant multiples of each other, $y = C_{1} y_{1} + C_{2} y_{2}$ is the general solution, with the two constants an order 2 equation needs.

The trial $y = e^{rx}$ . An exponential keeps its shape when differentiated, so try $y = e^{rx}$ : $y' = re^{rx}$ , $y'' = r^2e^{rx}$ , and

ay'' + by' + cy = (ar^2 + br + c)\,e^{rx} .

Since $e^{rx} \ne 0$ , $y = e^{rx}$ is a solution exactly when $a r^{2} + b r + c = 0$ . The discriminant $\Delta = b^2 - 4ac$ gives three cases.

Important

roots of $a r^{2} + b r + c = 0$	general solution
$\Delta \gt 0$ : two real roots $r_1 \ne r_2$	$y = C_1e^{r_1x} + C_2e^{r_2x}$
$\Delta = 0$ : a double root $r = -\frac{b}{2a}$	$y = (C_1 + C_2x)\,e^{rx}$
$\Delta \lt 0$ : complex roots $\alpha \pm i\beta$	$y = e^{\alpha x}\left(C_1\cos\beta x + C_2\sin\beta x\right)$

Why $xe^{rx}$ for a double root: with $y = xe^{rx}$ , $y' = (1 + rx)e^{rx}$ and $y'' = (2r + r^2x)e^{rx}$ , so $ay'' + by' + cy = \left[x(ar^2 + br + c) + (2ar + b)\right]e^{rx}$ . Both brackets are 0, because $r$ is a root and $r = -\frac{b}{2a}$ .

Why sines and cosines for complex roots: by Euler's formula from Unit 1, $e^{(\alpha + i\beta)x} = e^{\alpha x}(\cos\beta x + i\sin\beta x)$ . Its real and imaginary parts are each real solutions.

Worked example 1: distinct real roots

Solve $y^{''} - y^{'} - 6 y = 0$ with $y (0) = 1$ and $y^{'} (0) = 8$ .

Solution $r^{2} - r - 6 = (r - 3) (r + 2) = 0$ , so $y = C_1e^{3x} + C_2e^{-2x}$ and $y' = 3C_1e^{3x} - 2C_2e^{-2x}$ . The conditions give

C_1 + C_2 = 1, \qquad 3C_1 - 2C_2 = 8 .

Adding twice the first to the second: $5 C_{1} = 10$ , so $C_{1} = 2$ , $C_{2} = - 1$ , and $y = 2e^{3x} - e^{-2x}$ .

Worked example 2: a double root and complex roots

Solve (a) $y^{''} + 6 y^{'} + 9 y = 0$ , $y (0) = 2$ , $y^{'} (0) = - 5$ (b) $y^{''} + 2 y^{'} + 5 y = 0$ , $y (0) = 0$ , $y^{'} (0) = 4$ .

Solution

$(r + 3)^{2} = 0$ : double root $- 3$ , so $y = (C_1 + C_2x)e^{-3x}$ . From $y (0) = 2$ , $C_{1} = 2$ . Then $y' = \left[C_2 - 3(C_1 + C_2x)\right]e^{-3x}$ , so $y^{'} (0) = C_{2} - 6 = - 5$ and $C_{2} = 1$ : $y = (2 + x)e^{-3x}$ .
$r = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$ , so $\alpha = -1$ , $\beta = 2$ and $y = e^{-x}(C_1\cos 2x + C_2\sin 2x)$ . From $y (0) = 0$ , $C_{1} = 0$ . Then $y' = C_2e^{-x}(2\cos 2x - \sin 2x)$ , so $y^{'} (0) = 2 C_{2} = 4$ and $C_{2} = 2$ : $y = 2e^{-x}\sin 2x$ , an oscillation that dies away.

The roots of the characteristic equation decide the shape of every solution. Here each curve starts at y = 1 at rest, like a car's spring pressed down and released.

In mechanics the three cases have names. Two real roots: overdamped, a slow creep back. A double root: critically damped, the quickest return without overshooting, which is what a car's shock absorbers aim for. Complex roots: underdamped, a decaying oscillation. With no damping at all, $b = 0$ , the roots are $\pm i\beta$ and the solution $C_1\cos\beta x + C_2\sin\beta x$ is simple harmonic motion, as studied in physics.

Warning

A double root needs the factor $x$ . Writing $y = C_1e^{-3x} + C_2e^{-3x}$ is really $(C_1 + C_2)e^{-3x}$ : one constant, not two, and it cannot meet two initial conditions. The second solution is $xe^{-3x}$ .

Tip

The general solution can be read straight off the roots, so the real work is the quadratic. Check your roots: their sum must be $-\frac{b}{a}$ and their product $\frac{c}{a}$ . For $r^{2} + 2 r + 5$ : $(- 1 + 2 i) + (- 1 - 2 i) = - 2$ ✓ and $(- 1)^{2} + 2^{2} = 5$ ✓.

MIT OpenCourseWare: Unforced Damped Motion, with Gilbert Strang (Learn Differential Equations: Up Close). Watch how the roots of the characteristic equation decide whether the motion is overdamped, critically damped or underdamped, as in the graph above.

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Practice

Questions 6 to 8 work backwards and are harder.

Write the characteristic equation and the general solution of (a) $y^{''} - 5 y^{'} + 6 y = 0$ (b) $y^{''} - 4 y = 0$ (c) $y^{''} + 4 y^{'} + 4 y = 0$ (d) $y^{''} + 16 y = 0$ (e) $y^{''} - 2 y^{'} + 10 y = 0$ .
Solve $y^{''} - 3 y^{'} = 0$ with $y (0) = 2$ and $y^{'} (0) = 3$ .
Solve $y^{''} - 2 y^{'} + y = 0$ with $y (0) = 1$ and $y^{'} (0) = 3$ .
Solve $y^{''} + 4 y = 0$ with $y (0) = 3$ and $y^{'} (0) = - 4$ . What is the amplitude of the motion?
Find the general solution of $4 y^{''} + 4 y^{'} + 5 y = 0$ .
The general solution of $y^{''} + p y^{'} + q y = 0$ is $y = (C_1 + C_2x)e^{5x}$ . Find $p$ and $q$ .
Find a differential equation with general solution $y = e^{2x}(C_1\cos 3x + C_2\sin 3x)$ .
A mass on a spring obeys $x^{''} + 2 k x^{'} + 9 x = 0$ , with $k \ge 0$ . For which values of $k$ is the motion (a) oscillatory (b) critically damped (c) overdamped?

Show answersHide answers

(a) $r = 2, 3$ : $C_1e^{2x} + C_2e^{3x}$ (b) $r = \pm 2$ : $C_1e^{2x} + C_2e^{-2x}$ (c) $r = - 2$ (double): $(C_1 + C_2x)e^{-2x}$ (d) $r = \pm 4i$ : $C_1\cos 4x + C_2\sin 4x$ (e) $r = 1 \pm 3i$ : $e^x(C_1\cos 3x + C_2\sin 3x)$
$r = 0, 3$ : $y = C_1 + C_2e^{3x}$ ; $3 C_{2} = 3$ and $C_{1} + C_{2} = 2$ , so $y = 1 + e^{3x}$ .
$y = (C_{1} + C_{2} x) e^{x}$ ; $C_{1} = 1$ , $C_{1} + C_{2} = 3$ : $y = (1 + 2 x) e^{x}$ .
$y = 3\cos 2x - 2\sin 2x$ ; amplitude $\sqrt{3^2 + 2^2} = \sqrt{13} \approx 3.61$ .
$r = -\frac{1}{2} \pm i$ : $y = e^{-\frac{x}{2}}(C_1\cos x + C_2\sin x)$
Double root 5: $(r - 5)^{2} = r^{2} - 10 r + 25$ , so $p = - 10$ , $q = 25$ .
Roots $2 \pm 3i$ : $r^{2} - 4 r + 13 = 0$ , so $y^{''} - 4 y^{'} + 13 y = 0$ .
$\Delta = 4k^2 - 36$ : (a) $0 \le k \lt 3$ (b) $k = 3$ (c) $k \gt 3$

Sources and credits

Written by Ganzaa: the explanations, the derivations, the graph (drawn from computed solutions), the worked examples, the practice items and the quiz questions. Every solution was checked by substituting it back into its equation numerically.
Video: Unforced Damped Motion, Gilbert Strang, MIT OpenCourseWare, embedded from YouTube.

1Springs, shock absorbers and the second derivative

2Where the characteristic equation comes from

Springs, shock absorbers and the second derivative

Where the characteristic equation comes from