Integrals of a Differential Equation
DEFINITION. Let f(x, y, yv) = 0 be a differential equation. A real valued C1–function u(x, y) is
said to be an integral of the differential equation if a solution y = (x) of the differential equation
defined on the interval [a, b], then
x A u(x, (x))
is constant on the interval [a, b].
DEFINITION. If u(x, y) is an integral of the differential equation f(x, y, yv) = 0, the the contour
lines u(x, y) = c are called integral curves of the differential equation.
DEFINITION. A normal curve family is a family of curves defined by the equations y = f(x, C)
with f a C1 real valued function on a domain in ½2. A regular curve family in a domain D is a
family of curves such that one and only one curve goes through each point of D, each curve has a
tangent at each point, and the tangent direction is a continuous function of position.
THEOREM (Continuation Theorem). Let f(x, y) be a C1 function on an open region
of the
xy–plane. Given any point (x0, y0) D
, the differential equation
yv = f(x, y)
has a unique solution y = (x) satisfying the initial condition
(x0) = y0
and the solution is defined for an interval x0 ) x < b (where b may be equal to ') such that if b <
', either (x, (x)) approaches the boundary of
or such that (x) is unbounded as x A b.
EXAMPLE. Find the integrals of the differential equation
2xdx + 2ydy = 0.
SOLUTION. This an exact differential equation and its solution is given by
x2 + y2 = C.
If (x) is a solution of the DE on an interval [a, b], then
2x + 2 (x) v(x) = 0.
Then the function
x A x2 + (x)2
has derivative equal to 0 on [a, b] and thus is constant on [a, b]. By definition the function
u(x, y) = x2 + y2
is an integral for the 2xdx + 2ydy = 0.
The DE is actually
x
dy
=- .
y
dx
So the form of the DE is
Integrals of a Differential Equation page 1
yv = f(x, y)
with
x
f(x, y) = - .
y
The function f(x, y) is defined everywhere except for y = 0, i.e.,
Domain f = {(x, y) ° y & 0}.
Now take the specific solution of the implicit equation
y 2 + x2 = 4
with constant equal to 4 as
(x) = 4 - x2 .
The corresponds, for example, to an initial condition
(x0, y0) = (0, 2).
Then the solution (x) gets extended for all x until
(x, 4 - x2 )
runs out of the domain of f(x, y). So the solution (x) is defined on (-2, 2). We leave out the
endpoints since the derivative of (x) at x = ±2 is not defined or since (±2, 0) is not in Domain f.
QUESTION: We know that
(x(t), y(t)) = (2 cos t, 2 sin t)
(- ' < t < ')
satisfy
x2 + y2 = 4.
How is the curve
a(t) = (2 cos t, 2 sin t)
related to the integral curves. We have that
av(t) = (- 2 sin t, 2 cos t).
Then
a(t) = (x(t), y(t))
is the solution of the system of equations
dx
dy
= - y,
= x.
dt
dt
Suppose that xv(t0) & 0. Then we can find the inverse function x- 1 for x(t) in some neighborhood
P = (x0 - b, x0 + b)
around x(t0) = x0. Let
(u) = yux-1(u).
Then
d(yux-1)
yv(x-1(u)) x(x-1(u))
dx-1
u
-1
(u)=
(u)= yv(x (u)) (u)=
==.
du
du
du
xv(x-1(u)) y(x-1(u)) (u)
d So the function (u) satisfies our original DE
x
dy
=y
dx
Integrals of a Differential Equation page 2
on P around points where the function t A x(t) can be inverted.
ANSWER TO QUESTION: The curves
a(t) = (2 cos t, 2 sin t) = (x(t), y(t))
are solution curves of the DE
x
dy
=y
dx
whenever x(t) can be inverted (and also when y(t) & 0).
PRACTICAL OUTCOME: Often the parametric form of the solution curves are displayed for a
DE yv = f(x, y) when strictly speaking places where the tangent is vertical has no meaning.
Note that solution curves for a DE form a regular curve family, i.e., one curve at most through
each point. In fact, if two curves go through one point, say the point (x0, y0), then we have that
have that the two solutions defining the two curves, say (x, (x)) and (x, e(x)), satisfy
(x0, (x0)) = (x0, y0) = (x0, e(x0)).
By the uniqueness condition,
(x) = e(x)
over the largest interval (cf. Continuation Theorem) for which the solutions are defined.
Integrals of a Differential Equation page 3