Solutions of Ordinary Differential Equations
Consider the n-th order ODE F(x, y, y!, y!!,..., y( n ) ) = 0 (1).
Definition: Let g(x) be a real-valued function defined on a interval I, having the n-th
derivative for all x in I.
g(x) is called an explicit solution of the equation (1) on the interval I, if:
1) F(x,g(x), g!(x), g!!(x),...,g( n ) (x)) is defined for all x in I
2) F(x,g(x), g!(x), g!!(x),...,g( n ) (x)) = 0 for all x in I
Example:
1) Verify that the function g(x) = e2x is an explicit solution of the equation
F(x,y,y’,y’’) = y’’ + y’ – 6y = 0
We have g’(x) = 2e2x and g’’(x) = 4e2x, when we substitute in the equation, we get
F(x, g(x), g’(x), g’’(x))= 4e2x + 2e2x – 6 e2x = 0 and defined for all x real.
2
x3
2) Verify that the function h(x) =
defined on the interval (0, 3) is an explicit solution
of the equation F(x,y,y’) = 3xy’ – 2y = 0.
1
2 !
Since g’(x) = x 3 is defined on the interval (0, 3) and
3
1
2
2
2
2 !
F(x, g(x), g’(x)) = 3x x 3 ! 2x 3 = 2x 3 ! 2x 3 = 0 and defined for all x in (0, 3)
3
Definition: A relation H(x,y) = 0 is called an implicit solution of the ODE (1) if this
relation produces at least one real-valued function g(x) defined on the interval I, such that
g(x) is an explicit solution of (1) on I.
Example:
1) The relation x2 + y2 – 4 = 0 is an implicit solution of the equation
F(x,y,y’) = x + yy’= 0
on the interval (-2, 2).
The relation produces the functions h(x) = y =
4 ! x 2 and also the function
k(x) = y = - 4 ! x 2 both defined on (-2, 2).
"2x
Since h!(x) =
, then
2
2 4"x
!x 2
F(x,h(x),h’(x)) = x + 4 ! x 2
= 0 and defined on (-2, 2).
4 ! x2
2) The relation x2 + y2 + 4 = 0 is NOT an implicit solution of the equation
F(x,y,y’) = x + yy’= 0
on the interval (-2,2).
The relation does not produce a real-valued function on the interval, solving for y we get
p(x) = y = !4 ! x 2 and this function is undefined since - 4 – x2 < 0.
If we differentiate the relation implicitly, we get 2x + 2yy’ = 0 or x + yy’ = 0, that is the
equation we want to solve. Then, we say that the relation x2 + y2 + 4 = 0 is a formal
solution of the differential equation.
Remark: A first order ODE can be given by the expression
dy
= f(x, y)
dx
Geometric interpretation of the first order ODE
dy
A first order ODE
= f(x, y) associates with each point (x0, y0) in a region D ! ! 2 a
dx
dy
direction m =
= f(x 0 , y 0 ) . The direction at each point (x0, y0) is the slope of the
dx
tangent line to a curve, with equation g(x,y) = c, passing through the point.
The region D with the direction at each point creates what is called a direction field.
Solving the differential equation means to find curves whose tangent lines at the point
dy
(x0, y0) has slope m =
= f(x 0 , y 0 ) .
dx
Example:
dy
The ODE
= 2x creates a directional field in ! 2 . Every curve y = x2 + c where c is
dx
an arbitrary constant, has a tangent line at the point (x,y) with slope m = 2x.
Remark:
We saw in the above example that functions of the form hc(x) = x2 + c where c is any
dy
constant, are solutions of the ODE
= 2x . The constant c is called a parameter.
dx
Then, hc(x) = x2 + c is a one-parameter family of solutions of the ODE.
Every first order ODE has a one-parameter family of solutions.
Initial Value Problem
Let’s consider the following problem:
Find a function k(x) that is a solution of the ODE
dy
= 2x , such that k(1) = 4.
dx
This means that:
1- k(x) must satisfy the ODE, so k’(x) = 2x for all x
2- k(1) = 4
This is called an initial value problem (I.V.P.) and it is denoted by:
! dy
# = 2x
" dx
#$ y(1) = 4
dy
= 2x has a one-parameter family of solutions hc(x) = x2 + c where c
dx
is an arbitrary constant, imposing the condition x =1 then y = 1, we get 4 = (1)2 + c or
c = 3.
Therefore, k(x) = x2 + 3 satisfies the I.V.P..
Since the ODE
Remark:
For any first order ODE, we denote an I.V.P. by
! dy
# = f(x, y)
" dx
#$ y(x 0 ) = y 0
The I.V.P. can be extended to ODE of higher order.
Consider the problem
" d2y
$ 2 +y=0
$$ dx
# y(0) = !1
$ y '(0) = 0
$
$%
We have to find a solution h(x) such that satisfies the ODE and moreover h(0) = -1 and
h’(0) = 0.
For ODE of order 2 or higher there is another type of problem called boundary-value
problem and it is given by
" d2y
$ 2 +y=0
$$ dx
# y(0) = 1
$ y(!) = 0
$
$%
The conditions relate to the two different values of x, 1 and π.
Theorem: Basic Existence and Uniqueness Theorem
Given the I.V.P.
! dy
# = f(x, y)
" dx
#$ y(x 0 ) = y 0
!f
(x, y) are continuous in a rectangle
If the function f(x,y) and the partial derivative
!y
R centered at (x0,y0), R = {(x,y): x0- h < x < x0 + h, y0- k < y < y0 + k}
Then, there is a function p(x) defined on the interval (x0 – h, x0 + h) that is the unique
solution of the I.V.P.
Examples:
1) Consider the I.V.P.
x
" dy
$ =!
y
# dx
$ y(3) = 4
%
The relation x2 + y2 = c2 is a one-parameter family of solution of the ODE
dy
x
=! .
dx
y
The condition y(3) = 4 implies 32 + 42 = 25 = c2
Solving for y we get to functions:
h(x) =
25 ! x 2 and k(x) = - 25 ! x 2 both defined on the interval [-5,5].
Since h(3) = 4 and k(3) = -4, the only solution of the I.V.P. is h(x) = 25 ! x 2 .
x
!f
x
Notice that f(x,y) = ! and
= 2 are continuous on a rectangle centered at (3,4) and
y
!y y
does not include the x-axis.
2) Consider the I.V.P.
x
" dy
$ =!
y
# dx
$ y(1) = 0
%
x
"f
x
Since f(x, y) = ! !!and!! =
are NOT continuous along the x-axis, then we cannot
y
"y
y
!f
create a rectangle centered at (1,0) where f and
are continuous. The theorem fails and
!y
the I.V.P. has two solutions h(x) = 1 ! x 2 and k(x) = - 1 ! x 2
3) Consider the I.V.P.
! dy
# =2 y
" dx
#$ y(0) = 0
!f
2
is NOT defined at (0,0), then we cannot create a rectangle centered at
=
!y 2 y
!f
(0,0) where both f and
are continuous, therefore the theorem fails and the I.V.P. has
!y
two different solutions, h(x) = x2 and k(x) = 0.
Since