2
1.2.
First order ODEs.
The most general form is
F
(
x,y,y
′
) = 0
,
where
•
F
is a given function of 3 real variables;
•
x
is an independent variable;
•
y
is the unknown function
y
=
y
(
x
);
•
y
′
=
dy/dx
is the usual (
ordinary
) derivative.
Any diﬀerentiable function
y
(
x
) that satisﬁes the equation (i.e. if we substitute
y
(
x
) for
y
, then the equation becomes an identity) is called a
solution
.
Given two numbers
x
0
,y
0
, the corresponding
initial value problem
(”IVP”) is to
ﬁnd a solution
y
(
x
) such that
y
(
x
0
) =
y
0
.
Important
: we always assume that a solution is deﬁned on some open
interval
of
the real line; the interval can be inﬁnite, e.g. (
−∞
,
+
∞
).
Example
: we don’t consider the function
y
(
x
) =
{
log(
−
x
)
,
x <
0
,
log
x
+
C,
x >
0
,
which is deﬁned on the union of two intervals (
−∞
,
0)
∪
(0
,
+
∞
), to be a solution of
the equation
y
′
= 1
/x.
Otherwise, the IVP
y
(
−
1) = 0 would have inﬁnitely many
solutions.
A solution is
maximal
if it can not be extended to a solution on a larger interval.
1.3.
Higher order ODEs.
The general form of a 2-nd order ODE is
F
(
x,y,y
′
,y
′
,y
′′
) = 0
,
and solutions are
twice
diﬀerentiable functions (deﬁned on some intervals) satisfying
the equation.
Solving for the second derivative, we obtain Newton’s equation with 1 degree of
freedom (the motion of a particle
m
= 1 on the real line):
¨
x
=
f
(
t,x,
˙
x
)
.
[Here we have changed the notation: now time
t
is an independent variable, the
position
x
=
x
(
t
) is the unknown function, and dot means time derivative, so ¨
x
is
acceleration. The RHS is the ”force”; e.g.
f
= 0 for a free particle,
f
=
−
kx
for a
Hooke’s law spring, etc.]
Example:
the general solution of Eq. ¨
x
= 0 is
x
(
t
) =
C
1
+
C
2
t,
and the IVP
x
(
t
0
) =
x
0
,
˙
x
(
t
0
) = ˙
x
0
,
has a unique solution on
R
.