5Second-order differential equations
IA Differential Equations
5.2 Particular integrals
We now consider equations of the form
ay
00
+
by
0
+
cy
=
f
(
x
). We will come up
with several ways to find a particular integral.
5.2.1 Guessing
If the forcing terms are simple, we can easily “guess” the form of the particular
integral, as we’ve previously done for first-order equations.
f(x) y
p
(x)
e
mx
Ae
mx
sin kx
A sin kx + B cos kx
cos kx
polynomial p
n
(x) q
n
(x) = a
n
x
n
+ ··· + a
1
x + a
0
It is important to remember that the equation is linear, so we can superpose
solutions and consider each forcing term separately.
Example. Consider
y
00
−
5
y
0
+ 6
y
= 2
x
+
e
4x
. To obtain the forcing term 2
x
,
we need a first order polynomial
ax
+
b
, and to get
e
4x
we need
ce
4x
. Thus we
can guess
y
p
= ax + b + ce
4x
y
0
p
= a + 4ce
4x
y
00
p
= 16ce
4x
Substituting in, we get
16ce
4x
− 5(a + 4ce
4x
) + 6(ax + b + ce
4x
) = 2x + e
4x
Comparing coefficients of similar functions, we have
16c − 20c + 6c = 1 ⇒ c =
1
2
6a = 2 ⇒ a =
1
3
−5a + 6b = 0 ⇒ b =
5
18
Since the complementary function is
y
c
=
Ae
3x
+
Be
2x
, the general solution is
y = Ae
3x
+ Be
2x
+
1
2
e
4x
+
1
3
x +
5
18
.
Note that any boundary condition to determine
A
and
B
must be applied to
the full solution, not the complementary function
5.2.2 Resonance
Consider
¨y
+
ω
2
0
y
=
sin ω
0
t
. The complementary solution is
y
c
=
A sin ω
0
t
+
B cos w
0
t
. We notice that the forcing is itself a complementary function. So
if we guess a particular integral
y
p
=
C sin ω
0
t
+
D cos ω
0
t
, we’ll simply find
¨y
p
+ ω
2
0
y
p
= 0, so we can’t balance the forcing.
This is an example of a simple harmonic oscillator being forced at its natural
frequency.
We can detune our forcing away from the natural frequency, and consider
¨y + ω
2
0
y = sin ωt with ω 6= ω
0
. Try
y
p
= C(sin ωt − sin ω
0
t).
We have
¨y
p
= C(−ω
2
sin ωt + ω
2
0
sin ω
0
t).
Substituting into the differential equation, we have C(ω
2
0
− ω
2
) = 1. Then
y
p
=
sin ωt − sin ω
0
t
ω
2
0
− ω
2
.
We can simplify this to
y
p
=
2
ω
2
0
− ω
2
cos
ω
0
+ ω
2
t
sin
ω −ω
0
2
t
We let ω
0
− ω = ∆ω. Then
y
p
=
−2
(2ω + ∆ω)∆ω
cos
ω +
∆ω
2
t
sin
∆ω
2
t
.
t
y
p
y
p
sin
∆ω
2
t
−sin
∆ω
2
t
O
1
∆ω
This oscillation in the amplitude of the
cos
wave is known as beating. This
happens when the forcing frequency is close to the natural frequency. The
wavelength of the
sin
function has order
O
(
1
∆ω
) and
cos
has wavelength
O
(
1
ω
0
).
As ∆
ω →
0, the wavelength of the beating envelope
→ ∞
and we just have the
initial linear growth.
Mathematically, since sin θ ≈ θ as θ → 0, as ∆ω → 0, we have
y
p
→
−t
2ω
0
cos ω
0
t.
In general, if the forcing is a linear combination of complementary functions,
then the particular integral is proportional to
t
(the independent variable) times
the non-resonant guess.
5.2.3 Variation of parameters
So far, we have been finding particular integrals by guessing. Here we are going
to come up with a method that can systematically help us find the particular
integral. Of course, this is substantially more complicated than guessing. So if
the form of the particular integral is obvious, we should go for guessing.
Suppose we have a second order differential equation
y
00
+ p(x)y
0
+ q(x)y = f(x).
We then know any particular solution vector can be written in the form
Y(x) =
y(x)
y
0
(x)
,
and our job is to find one solution vector that satisfies the equation. We
presuppose such a solution actually exists, and we will try to find out what it is.
The trick is to pick a convenient basis for this space. Let
y
1
(
x
) and
y
2
(
x
) be
linearly independent complementary functions of the ODE. Then for each
x
, the
solution vectors Y
1
(
x
) = (
y
1
(
x
)
, y
0
1
(
x
)) and Y
2
(
x
) = (
y
2
(
x
)
, y
0
2
(
x
)) form a basis
of the solution space. So for each particular
x
, we can find some constants
u, v
(depending on x) such that the following equality holds:
Y(x) = u(x)Y
1
(x) + v(x)Y
2
(x),
since Y
1
and Y
2
are a basis.
Component-wise, we have
y
p
= uy
1
+ vy
2
(a)
y
0
p
= uy
0
1
+ vy
0
2
(b)
Differentiating the second equation, we obtain
y
00
p
= (uy
00
1
+ u
0
y
0
1
) + (vy
00
2
+ v
0
y
0
2
) (c)
If we consider (c) + p(b) + q(a), we have y
0
1
u
0
+ y
0
2
v
0
= f.
Now note that we derived the equation of
y
0
p
from the vector equation. This
must be equal to what we get if we differentiate (a). By (a)
0
- (b), we obtain
y
1
u
0
+
y
2
v
0
= 0. Now we have two simultaneous equations for
u
0
and
v
0
, which
we should be able to solve.
We can, for example, write them in matrix form as
y
1
y
2
y
0
1
y
0
2
u
0
v
0
=
0
f
Inverting the left matrix, we have
u
0
v
0
=
1
W
y
0
2
−y
2
−y
0
1
y
1
0
f
So u
0
= −
y
2
W
f and v
0
=
y
1
W
f.
Example.
y
00
+ 4
y
=
sin
2
x
. We know that
y
1
=
sin
2
x
and
y
2
=
cos
2
x
.
W = −2. We write
y
p
= u sin 2x + v cos 2x.
Using the formulae above, we obtain
u
0
=
cos 2x sin 2x
2
=
sin 4x
4
, v
0
=
−sin
2
2x
2
=
cos 4x − 1
4
So
u = −
cos 4x
16
, v =
sin 4x
16
−
x
4
Therefore
y
p
=
1
16
(−cos 4x sin 2x + sin 4x cos 2x −
x
4
cos 2x) =
1
16
sin 2x −
x
4
cos 2x
Note that
−
1
4
x cos
2
x
is what we found previously by detuning, and
1
16
sin
2
x
is
a complementary function, so the results agree.
It is generally not a good idea to remember the exact formula for the results
we’ve obtained above. Instead, whenever faced with such questions, you should
be able to re-derive the results instead.