The University of Sydney
School of Mathematics and Statistics
Lectures Week 5 – Second-Order Linear Equations
MATH1023: Multivariable Calculus and Modelling Semester 1, 2023
1. Solve 2nd-Order, Linear, Constant Coefficient, Homogeneous DEs.
2. Calculate the Characteristic equation.
3. Find the general solution using the roots of the Characteristic equation.
4. Study the equation of the Simple Harmonic Oscillator
5. Solve the equation of the Damped Harmonic Oscillator and identify the three cases:
supercritical damping, subcritical damping, and critical damping.
2nd-Order, Linear, Constant Coefficient, Homogeneous DEs
General Form:
d2y
dx2
+ a
dy
dx
+ by = 0 with a, b constants.
How would one solve this equation?
1
Solution of the characteristic equation
m2 + am+ b = 0 =⇒ m1, m2 = 1
2
(
− a±
√
a2 − 4b
)
If a2 6= 4b then m1 6= m2 and we have two solutions y1(x) = C em1x and y2(x) = D em2x,
where C, D are constants. Since the sum of two solutions is also a solution, we yield the
general solution:
y = Cem1x +Dem2x
From here, we may look at the nature of the solutions of the DE.
• Case 1 a2 > 4b
Two distinct real roots y = Cem1x +Dem2x
– If both m1 and m2 are positive
– If both m1 and m2 are negative
– If m1 and m2 are of opposite sign
2
• Case 2 a2 < 4b
Complex conjugate roots
m1 = −a/2 + k i, m2 = −a/2− k i where k =
√
4b− a2/2)
The general solution can be written in complex form as
y = C em1x +D em2x
or
y = e−ax/2
(
Ceikx +De−ikx
)
or
y = e−ax/2(Cˆ cos kx+ Dˆ sin kx)
(using Euler’s formulas).
– a < 0 =⇒
– a > 0 =⇒
– a = 0 =⇒
3
• Example 1 Solve the DE y′′ + y′ − 20y = 0 with initial conditions y(0) = 1 and
y(0) = 1.
• Example 2 Solve the DE y′′ + 4y′ + 7y = 0 with initial conditions y(0) = 0 and
y′(0) = 3.
Summary of Nature of Solutions
m1, m2 =
1
2
(
− a±
√
a2 − 4b
)
• Case 1 a2 > 4b Two distinct real roots m1 6= m2
• Case 1.1 m1, m2 > 0 =⇒ Two growing exponentials
• Case 1.2 m1, m2 < 0 =⇒ Two decaying exponentials
• Case 1.3 m1, m2 =⇒ Opposite signs - One Grows, the other Decays
• Case 2 a2 < 4b Two complex roots
• Case 2.1 a < 0 =⇒ The solution curve oscillates between two growing exponential curves
• Case 2.2 a > 0 =⇒ The solution curve oscillates between two decaying exponential curves
• Case 2.3 a = 0 =⇒ Purely sinusoidal
• Case 3 a2 = 4b One real double root (Special case)
• Case 3 a2 = 4b,
One double root m1 = m2 = −a/2, one solution y = e−ax/2; we have to find a second
solution.
The general solution to the original problem derives as
y = (Ax+B)e−ax/2
A, B arbitrary constants.
Example 1 Solve the DE y′′ − 2y′ + y = 0 with boundary conditions y(0) = 0 and
y(1) = 1.
Example 2 Solve the DE y′′+3y′+2y = 0 with initial conditions y(0) = 1 and y′(0) = 1.
Example 3 Solve the DE y′′ + 9y = 0 with initial conditions y(0) = 1 and y′(0) = 0.
4
The Harmonic Oscillator
• Many systems oscillate or vary periodically.
• Examples: musical instruments, ocean waves, structures such as bridges, Boom-Bust
economic cycles, the Earth’s surface during an earthquake, love, and so on!
• The equation that describes oscillatory phenomena is called the Harmonic Oscillator
equation: y¨ + 2 γ y˙ + ω20 y = 0 where y˙ = dy/dt.
There are two main cases:
• When γ = 0 , it represents simple harmonic motion. No damping is present.
• When γ > 0 , it is a damped harmonic oscillator without forcing term.
Simple Harmonic Motion (SHM) (γ = 0)
The equation is y¨ + ω20 y = 0 and the general solution is:
y = C cosω0t+D sinω0t (C, D = Arbitrary constants)
Alternatively, we may write the general solution in the form
y = A cos(ω0t+ φ) (A, φ = Arbitrary constants)
A is called the amplitude, ω0 is called the frequency (measured in radians/second), φ is
called the phase of the oscillation, 0 ≤ φ ≤ 2pi, and T = 2pi/ω0 is defined as the period of
the oscillator.
The Damped Harmonic Oscillator
When γ 6= 0, there are three possible kinds of physical behaviour:
• Case 1 γ2 > ω20 =⇒ Supercritical damping
Solution is y = e−γt
(
Ceκt +De−κt
) (
κ =
√|γ2 − ω20|)
Alternatively y = Ce−(γ−κ)t +De−(γ+κ)t
The exponents are both real and negative for t > 0,
since κ =
√
γ2 − ω20 < γ . Thus both terms decay exponentially as t→∞.
5
• Case 2 γ2 < ω20 =⇒ Subcritical damping
Solution is y = e−γt (C cosκt+D sinκt)
Alternatively y = ae−γt cos(κt+ φ)
The cosine term produces an oscillation. However, note the amplitude is modulated
by e−γt . Thus, the solution decays exponentially as t → ∞, and the oscillation dies
away.
• Case 3 γ2 = ω20 =⇒ Critical damping
Solution is y = e−γ t (C +Dt)
The term t e−γ t increases initially but will eventually decay.
• Thus in all three cases, positive damping γ > 0 causes the displacement to die
away as t→∞