5Second-order differential equations
IA Differential Equations
5.1 Constant coefficients
The general form of an equation with constant coefficients is
ay
00
+ by
0
+ cy = f(x).
We solve this in two steps:
(i)
Find the complementary functions which satisfy the homogeneous equation
ay
00
+ by
0
+ cy = 0.
(ii) Find a particular solution that satisfies the full equation.
5.1.1 Complementary functions
Recall that
e
λx
is an eigenfunction of the differential operator
d
dx
. Hence it is
also an eigenfunction of the second derivative
d
2
dx
2
=
d
dx
d
dx
.
If the complementary function has the form
y
c
=
e
λx
, then
y
0
c
=
λe
λx
and
y
00
c
= λ
2
e
λx
. Substituting into the differential equation gives
Definition (Characteristic equation). The characteristic equation of a (second-
order) differential equation ay
00
+ by
0
+ c = 0 is
aλ
2
+ bλ + c = 0.
In this case there are two solutions to the characteristic equation, giving (in
principle) two complementary functions y
1
= e
λ
1
x
and y
2
= e
λ
2
x
.
If
λ
1
and
λ
2
are distinct, then
y
1
and
y
2
are linearly independent and complete
— they form a basis of the solution space. The (most) general complementary
function is
y
c
= Ae
λ
1
x
+ Be
λ
2
x
.
Example.
y
00
−
5
y
0
+ 6
y
= 0. Try
y
=
e
λx
. The characteristic equation is
λ
2
− 5λ + 6 = 0. Then λ = 2 or 3. So the general solution is y = Ae
2x
+ Be
3x
.
Note that A and B can be complex constants.
Example (Simple harmonic motion).
y
00
+ 4
y
= 0. Try
y
=
e
λx
. The character-
istic equation is
λ
2
+ 4 = 0, with solutions
λ
=
±
2
i
. Then our general solution
is
y
=
Ae
2ix
+
Be
−2ix
. However, if this is in a case of simple harmonic motion
in physics, we want the function to be real (or look real). We can write
y = A(cos 2x + i sin 2x) + B(cos 2x −i sin 2x)
= (A + B) cos 2x + i(A − B) sin 2x
= α cos 2x + β sin 2x
where α = A + B and β = i(A − B), and α and β are independent constants.
In effect, we have changed the basis from {e
2ix
, e
−2ix
} to {cos 2x, sin 2x}.
Example (Degeneracy). y
00
− 4y
0
+ 4y = 0.
Try
y
=
e
λx
. We have
λ
2
−
4
λ
+ 4 = 0 and (
λ −
2)
2
= 0. So
λ
= 2 or 2. But
e
2x
and
e
2x
are clearly not linearly independent. We have only managed to find
one basis function of the solution space, but a second order equation has a 2
dimensional solution space. We need to find a second solution.
We can perform detuning. We can separate the two functions found above
from each other by considering
y
00
−
4
y
0
+ (4
− ε
2
)
y
= 0. This turns into the
equation we want to solve as
ε →
0. Try
y
=
e
λx
. We obtain
λ
2
−
4
λ
+ 4
− ε
2
.
The two roots are λ = 2 ± ε. Then
y = Ae
(2+ε)x
+ Be
(2−ε)x
= e
2x
[Ae
εx
+ Be
−εx
]
Taking the limit as ε → 0, we use the Taylor expansion of e
εx
to obtain
y = e
2x
[(A + B) + εx(A − B) + O(Aε
2
, Bε
2
)]
We let (
A
+
B
) =
α
and
ε
(
A − B
) =
β
. This is perfectly valid for any non-zero
ε. Then A =
1
2
(α +
β
ε
) and B =
1
2
(α −
β
ε
). So we have
y = e
2x
[α + βx + O(Aε
2
, Bε
2
)]
We know for any
ε
, we have a solution of this form. Now we turn the procedure
around. We fix some
α
and
β
. Then given any
ε
, we can find some constants
A
,
B
(depending on
ε
) such that the above holds. As we decrease the size of
ε
, we
have A, B = O(
1
ε
). So O(Aε
2
) = O(Bε
2
) = O(ε). So our solution becomes
y = e
2x
[α + βx + O(ε)]
→ e
2x
(α + βx)
In this way, we have derived two separate basis functions. In general, if
y
1
(
x
)
is a degenerate complementary function of a linear differential equation with
constant coefficients, then
y
2
(
x
) =
xy
1
(
x
) is an independent complementary
function.
5.1.2 Second complementary function
In general (i.e. if we don’t have constant coefficients), we can find a second com-
plementary function associated with a degenerate solution of the homogeneous
equation by looking for a solution in the form
y
2
(
x
) =
v
(
x
)
y
1
(
x
), where
y
1
(
x
) is
the degenerate solution we found.
Example. Consider
y
00
−
4
y
0
+ 4
y
= 0. We have
y
1
=
e
2x
. We try
y
2
=
ve
2x
.
Then
y
0
2
= (v
0
+ 2v)e
2x
y
00
2
= (v
00
+ 4v
0
+ 4v)e
2x
.
Substituting into the original equation gives
(v
00
+ 4v
0
+ 4v) − 4(v
0
+ 2v) + 4v = 0.
Simplifying, this tells us v
00
= 0, which forces v to be a linear function of x. So
y
2
= (Ax + B)e
2x
for some A, B ∈ R.
5.1.3 Phase space
If we are given a general nth order differential equation of the form
a
n
(x)y
(n)
+ a
n−1
y
(n−1)
+ ··· + a
1
(x)y
0
+ a
0
(x)y = f(x),
and we have a solution
y
, then we can plot a graph of
y
versus
x
, and see how
y
evolves with x.
However, one problem is that for such an equation, the solution is not just
determined by the initial condition
y
(
x
0
), but also
y
0
(
x
0
),
y
00
(
x
0
) etc. So if we
just have a snapshot of the value of
y
at a particular point
x
0
, we have completely
no idea how it would evolve in the future.
So how much information do we actually need? At any point
x
0
, if we
are given the first
n −
1 derivatives, i.e.
y
(
x
0
),
y
0
(
x
0
),
···
,
y
(n−1)
(
x
0
), we can
then get the
n
th derivative and also any higher derivatives from the differential
equation. This means that we know the Taylor series of
y
about
x
0
, and it follows
that the solution is uniquely determined by these conditions (note that it takes
considerably more extra work to actually prove rigorously that the solution is
uniquely determined by these initial conditions, but it can be done for sufficiently
sensible f, as will be done in IB Analysis II).
Thus we are led to consider the solution vector
Y(x) = (y(x), y
0
(x), ··· , y
n−1
(x)).
We say such a vector lies in the phase space, which is an
n
-dimensional space. So
for each
x
, we thus get a point Y(
x
) lying in the
n
-dimensional space. Moreover,
given any point in the phase space, if we view it as the initial conditions for our
differential equation, we get a unique trajectory in the phase space.
Example. Consider
y
00
+ 4
y
= 0. Suppose we have an initial condition of
y
1
(0) = 1
, y
0
1
(0) = 0. Then we can solve the equation to obtain
y
1
(
x
) =
cos
2
x
.
Thus the initial solution vector is Y
1
(0) = (1
,
0), and the trajectory as
x
varies is
given by Y
1
(
x
) = (
cos
2
x, −
2
sin
2
x
). Thus as
x
changes, we trace out an ellipse
in the clockwise direction:
y
y
0
Y
1
(x)
Another possible initial condition is
y
2
(0) = 0
, y
0
2
(0) = 2. In this case, we obtain
the solution y(x) = sin 2x, with a solution vector Y
2
(x) = (sin 2x, 2 cos 2x).
Note that as vectors, our two initial conditions (1
,
0) and (0
,
2) are indepen-
dent. Moreover, as
x
changes, the two solution vectors Y
1
(
x
)
,
Y
2
(
x
) remain
independent. This is an important observation that allows the method of varia-
tion of parameters later on.
In general, for a 2nd order equation, the phase space is a 2-dimensional space,
and we can take the two complementary functions Y
1
and Y
2
as basis vectors
for the phase space at each particular value of
x
. Of course, we need the two
solutions to be linearly independent.
Definition (Wronskian). Given a differential equation with solutions
y
1
, y
2
, the
Wronskian is the determinant
W (x) =
y
1
y
2
y
0
1
y
0
2
.
Definition (Independent solutions). Two solutions
y
1
(
x
) and
y
2
(
x
) are indepen-
dent solutions of the differential equation if and only if Y
1
and Y
2
are linearly
independent as vectors in the phase space for some
x
, i.e. iff the Wronskian is
non-zero for some x.
In our example, we have W (x) = 2 cos
2
2x + 2 sin
2
2x = 2 6= 0 for all x.
Example. In our earlier example, y
1
= e
2x
and y
2
= xe
2x
. We have
W =
e
2x
xe
2x
2e
2x
e
2x
+ 2xe
2x
= e
4x
(1 + 2x − 2x) = e
4x
6= 0.
In both cases, the Wronskian is never zero. Is it possible that it is zero for
some x while non-zero for others? The answer is no.
Theorem (Abel’s Theorem). Given an equation
y
00
+
p
(
x
)
y
0
+
q
(
x
)
y
= 0, either
W
= 0 for all
x
, or
W 6
= 0 for all
x
. i.e. iff two solutions are independent for
some particular x, then they are independent for all x.
Proof. If y
1
and y
2
are both solutions, then
y
2
(y
00
1
+ py
0
1
+ qy
1
) = 0
y
1
(y
00
2
+ py
0
2
+ qy
2
) = 0
Subtracting the two equations, we have
y
1
y
00
2
− y
2
y
00
1
+ p(y
1
y
0
2
− y
2
y
0
1
) = 0
Note that
W
=
y
1
y
0
2
−y
2
y
0
1
and
W
0
=
y
1
y
00
2
+
y
0
1
y
0
2
−
(
y
0
2
y
0
1
+
y
2
y
00
1
) =
y
1
y
00
2
−y
2
y
00
1
W
0
+ P (x)W = 0
W (x) = W
0
e
−
R
P dx
,
Where
W
0
= const. Since the exponential function is never zero, either
W
0
= 0,
in which case W = 0, or W
0
6= 0 and W 6= 0 for any value of x.
In general, any linear
n
th-order homogeneous differential equation can be
written in the form Y
0
+
A
Y = 0, a system of first-order equations. It can then
be shown that
W
0
+
tr
(
A
)
W
= 0, and
W
=
W
0
e
−
R
tr A dx
. So Abel’s theorem
holds.