4First-order differential equations
IA Differential Equations
4.2 Homogeneous linear ordinary differential equations
Before we start, we will first define the terms used in the title. These are useful
criteria for categorizing differential equations.
Definition (Linear differential equation). A differential equation is linear if the
dependent variable (y, y
0
, y
00
etc.) appears only linearly.
Definition (Homogeneous differential equation). A differential equation is
homogeneous if y = 0 is a solution.
Definition (Differential equation with constant coefficients). A differential
equation has constant coefficients if the independent variable
x
does not appear
explicitly.
Definition (First-order differential equation). A differential equation is first-
order if only first derivatives are involved.
Theorem. Any linear, homogeneous, ordinary differential equation with constant
coefficients has solutions of the form e
mx
.
This theorem is evident when we consider an example.
Example. Given 5
dy
dx
− 3y = 0. Try y = e
mx
. Then
dy
dx
= me
mx
5me
mx
− 3e
mx
= 0
Since this must hold for all values of
x
, there exists some value
x
for which
e
mx
6
= 0 and we can divide by
e
mx
(note in this case this justification is not
necessary, because
e
mx
is never equal to 0. However, we should justify as above
if we are dividing, say,
x
m
). Thus 5
m −
3 = 0 and
m
= 3
/
5. So
y
=
e
3x/5
is a
solution.
Because the equation is linear and homogeneous, any multiple of a solution
is also a solution. Therefore y = Ae
3x/5
is a solution for any value of A.
But is this the most general form of the solution? One can show that an
n
th
-order linear differential equation has
n
and only
n
independent solutions. So
y = Ae
3x/5
is indeed the most general solution.
We can determine A by applying a given boundary condition.
Discrete equations
Suppose that we are given the equation 5
y
0
−
3
y
= 0 with boundary condition
y
=
y
0
at
x
= 0. This gives a unique function
y
. We can approximate this by
considering discrete steps of length
h
between
x
n
and
x
n+1
. (Using the simple
Euler numerical scheme,) we can approximate the equation by
5
y
n+1
− y
n
h
− 3y
n
≈ 0.
Rearranging the terms, we have
y
n+1
≈
(1 +
3
5
h
)
y
n
. Applying the relation
successively, we have
y
n
=
1 +
3
5
h
y
n−1
=
1 +
3
5
h
1 +
3
5
h
y
n−2
=
1 +
3
5
h
n
y
0
For each given value of x, choose h = x/n, so
y
n
= y
0
1 +
3
5
(x/n)
n
.
Taking the limit as n → ∞, we have
y(x) = lim
n→∞
y
0
1 +
3x/5
n
n
= y
0
e
3x/5
,
in agreement with the solution of the differential equation.
x
y
solution of diff. eq.
1
discrete approximation
O
x
0
x
1
x
2
x
3
Series solution
We can also try to find a solution in the form of a Taylor Series
y
=
∞
P
n=0
a
n
x
n
.
We have y
0
=
P
a
n
nx
n−1
. Substituting these into our equation
5y
0
− 3y = 0
5(xy
0
) −3x(y) = 0
X
a
n
(5n −3x)x
n
= 0
Consider the coefficient of
x
n
: 5
na
n
−
3
a
n−1
= 0. Since this holds for all values
of
n
, when
n
= 0, we get 0
a
0
= 0. This tells that
a
0
is arbitrary. If
n >
0, then
a
n
=
3
5n
a
n−1
=
3
2
5
2
1
n(n −1)
a
n−2
= ··· =
3
5
n
1
n!
a
0
.
Therefore we have
y = a
0
∞
X
n=0
3x
5
n
1
n!
h
= a
0
e
3x/5
i
.