MA 18.03, R05
SECOND ORDER ODEs
1. A second order differential equation is an equation of the form
F (x, y, y 0 , y 00 ) = 0.
A solution of the differential equation is a function y = y(x) that satisfies the equation.
A differential equation has infinitely many solutions. They usually involve two undetermined constants.
2. An initial value problem consists of a differential equation and an initial conditions y(x0 ) =
y0 , y 0 (x0 ) = y00 . It has a unique solution, which has to be a function (provided F doesn’t
have problems at that point).
3. A boundary problem consists of a differential equation and conditions
y(x0 ) = y0 ,
y(x1 ) = y1 ;
y 0 (x0 ) = y00 ,
y(x1 ) = y1 ;
y(x0 ) = y0 ,
y 0 (x1 ) = y10 ;
y 0 (x0 ) = y00 ,
y 0 (x1 ) = y10 .
or
or
or
A. Homogeneous linear second order ODEs with constant coefficients
mx00 + bx0 + kx = 0,
m, b, k constants, m 6= 0
1. If the equation describes a physical spring–dashpot system, the coefficients m, b, k are
non-negative.
2. An initial value problem consists of a differential equation and an initial condition x(t0 ) =
A, x0 (t0 ) = B. How many solutions does it have?
3. The general solution is of the form x(t) = c1 x1 (t) + c2 x2 (t), where x1 and x2 are two
linearly independent solutions (none of them can be written as a constant multiple of the
other).
4. The characteristic polynomial of this equations is p(s) = ms2 + bs + k.
5. The exponential solutions of this equation are c1 er1 t and c2 er2 t , where r1 , r2 are the roots
(real or complex) of the characteristic polynomial and c1 , c2 are arbitrary constants. If
r1 = r2 = r, there is only one family of exponential solutions, namely cert .
How to solve:
1. Write down the characteristic equation ms2 + bs + k = 0.
2. Compute its discriminant ∆ = b2 − 4mk.
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MA 18.03, R05
3. There are three possible situations:
• overdamped: If ∆ > 0, the quadratic equation has two distinct real solutions, r1 and
r2 . Find them. (You might need to use the quadratic formula.) The general solution
of the differential equation is
x = C 1 e r1 t + C 2 e r2 t .
• critically damped: If ∆ = 0, the quadratic equation has only one real root r. Find it.
The general solution of the differential equation is
x = C1 ert + C2 tert .
• underdamped: If ∆ < 0, the quadratic equation does not have any real roots,
√ it has
|∆|
b
two complex conjugate roots r1,2 = α ± iβ, where α = − 2m and β = 2m . The
general solution of the differential equation is
x = C1 eαt cos(βt) + C2 eαt sin(βt)
or
x = Aeαt cos(βt − φ).
In the second expression, A is any positive real number and φ any number in [0, 2π).
4. If it is an initial value problem or a boundary problem, plug in the given values and solve
for C1 and C2 (or for A and φ). Don’t forget to take the derivative of x in the case of an
initial value problem (chain rule!).
B. Nonhomogeneous second order linear ODEs
mx00 + bx0 + kx = f (t).
We solve these by finding a particular solution xp of the given ODE and the solution xh to
the corresponding homogeneous equation mx00 + bx0 + kx = 0. The general solution is given by
x = xp + xh .
Note: xh contains all the undetermined constants (denotes an infinite family of functions),
while xp is one particular function.
We have two ways of finding xp , namely ERF and undetermined coefficients. They are both
explained below for linear equations of arbitrary order. Just apply them for your second order
ODE.
C. “Physical” notions
For a sinusoidal function C cos(ωt − φ)
the amplitude is |C|. Note that here C can be a complex number.
the angular frequency is ω.
the frequency is ω/(2π).
the period is P = 2π/ω.
For a damped sinusoidal Aert cos(ωd t − φ), we say that its pseudofrequency is ωd .
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MA 18.03, R05
Stability: a system described by a homogeneous second order equation
mx00 + bx0 + kx = 0
is stable if all solutions go to zero as t → ∞. This happens if the roots are real (maybe
repeated) and negative or if the roots are complex with the real part negative.
For a second order ODE with sinusoidal input
mx00 + bx0 + kx = B cos ωt
√
the natural frequency is k.
B>0
1
.
p(iω)
It has this name because the periodic solution to the above ODE is (as given by ERF)
the complex gain is the function ω 7→ W (iω) =
xp =
B cos ωt
= W (iω)B cos(ωt)
p(iw)
the amplitude gain is the absolute value of the complex gain |W (iω)| =
Think of it as
1
.
|p(iω)|
amplitude of the solution (output)
.
amplitude of the input
phase lag: write the periodic solution in the form xp = A cos(ωt − φ) with A > 0 and
0 ≤ φ < 2π. The phase lag is φ. It can also be computed as φ = − Arg(W (iω)) =
Arg(p(iω)).
the time lag is t0 > 0 such that the periodic solution xp can be written in the form
φ
xp = A cos(ω(t − t0 )), again with A > 0. It can also be computed as t0 = .
ω
pure resonance is achieved in case b = 0 (no damping) for a frequency ω if p(iω) = 0.
Then the equation has no periodic solution. (You need to apply ERF’ and the solution
will contain a t cos term.)
practical resonance is achieved in the case b 6= 0 (there is damping) for the values ω for
which the amplitude of the periodic solution is maximal. That is those ω for which
|p(iω)|2 = (k − mω 2 )2 + (bω)2 is minimal. Such an ω is called resonant frequency.
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MA 18.03, R05
LINEAR ODEs AND LINEAR OPERATORS
A. Linear operators
Notation
• D stands for the derivative of a function. For instance, if x is a function of t, Dx =
x0 = dx/dt. If f is a function of x, then Df = f 0 = df /dx.
• Dn stands for taking the n-th derivative. For instance D2 x = D(Dx) = D(x0 ) = x00 .
• I is the identity operator that leaves functions alone, so I(f ) = f.
• In general p(D) = an Dn + . . . + a1 D + a0 I applied to a function y means taking a
linear combination of y and its first n derivatives, namely
p(D)y = (an Dn + . . . + a1 D + a0 I)y = an y (n) + . . . + a1 y 0 + a0 y
Writing p(D)y = f (t) provides an handy shorthand for the linear ODE
an y (n) + . . . + a1 y 0 + a0 y = f (t)
Properties
• Linearity
p(D)(af + bg) = ap(D)f + bp(D)g
• Exponential shift formula
p(D)(ert u) = ert p(D + rI)u
Don’t forget that (D + rI)2 = D2 + 2rD + r2 I (and so forth).
B. Homogeneous linear ODEs with constant coefficients
an y (n) + . . . + a1 y 0 + a0 y = 0
This equation can be re-written as
p(D)y = 0
with
p(D) = an Dn + . . . + a1 D + a0 I.
1. Write down the characteristic polynomial p(s) = an sn + . . . + a1 s + a0 .
2. Find its n roots r1 , . . . , rn (counted with multiplicity) by solving an rn + . . . + a0 = 0.
3. The general solution is of the form x = c1 x1 + . . . + cn xn where each xj corresponds to the
root rj as follows.
• For each simple real root rj we obtain an exponential solution xj = erj t .
• For each pair of simple complex conjugate roots rj,j+1 = a ± ib we get the pair of
solutions xj = eat cos bt and xj+1 = eat sin bt.
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• A repeated real root rj with multiplicity m, should have m corresponding solutions.
Construct xj as above. The rest are txj , t2 xj , . . . , tm−1 xj .
• For repeated pairs of complex roots the same principle applies, except now you have
m pairs of corresponding solutions.
C. Non-homogeneous linear ODEs
an y (n) + . . . + a1 y 0 + a0 y = f (t)
This equation can be re-written as
p(D)y = f (t)
with
p(D) = an Dn + . . . + a1 D + a0 I.
We solve these by finding a particular solution xp of the given ODE and the solution xh to the
corresponding homogeneous equation an y (n) + . . . + a1 y 0 + a0 y = 0. The general solution is given
by
x = xp + xh .
Note: xh contains all the undetermined constants (denotes an infinite family of functions), while
xp is one particular function.
We have two ways of finding xp , detailed below. Recall that if f (t) = f1 (t) + f2 (t), look for
solutions xj (t) corresponding to fj (t), j = 1, 2, and then add them up to get xp = x1 + x2 . This
is called superposition.
Exponential response formulas: can be applied to an equation with exponential input
p(D)x = Aekt ,
where k is any complex number.
1. Write down the characteristic polynomial p(s).
2. Compute p(k). If it is nonzero, then xp =
Aekt
is a solution. (ERF)
p(k)
3. If p(k) = 0, compute p0 (k). If this in nonzero, then xp =
Atekt
is a solution. (ERF’)
p0 (k)
4. Keep going. The idea is to compute p(k), p0 (k), p00 (k), . . . until you find the first
nonzero guy amongst them. Say that happens to be p(m) (k). Then
xp =
is a solution.
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Atm ekt
p(m) (k)
MA 18.03, R05
Undetermined coefficients method: the idea is to look for a solution of the same general
form as the function f (t) on the RHS of our equation.
If f (t) is of the form
try
aert
Aert
a cos ωt
A cos ωt + B sin ωt
a sin ωt
A cos ωt + B sin ωt
an tn + . . . + a0
An t n + . . . + A0
(an tn + . . . + a0 )ert
(An tn + . . . + A0 )ert
(an tn + . . . + a0 ) cos ωt
(An tn + . . . + A0 )(B1 cos ωt + B2 sin ωt)
(an tn + . . . + a0 ) sin ωt
(An tn + . . . + A0 )(B1 cos ωt + B2 sin ωt)
(an tn + . . . + a0 )ert cos ωt
(An tn + . . . + A0 )ert (B1 cos ωt + B2 sin ωt)
(an tn + . . . + a0 )ert sin ωt
(An tn + . . . + A0 )ert (B1 cos ωt + B2 sin ωt)
If the form indicated on the right does not work, try multiplying it by t. If that doesn’t
work either, try multiplying it by t2 . Keep multiplying by higher and higher powers of t
until it works. But the power should never be higher than the order of the ODE. If you
passed this threshold, something is very, very wrong!
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