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) = e
2x
is an explicit solution of the equation
F(x,y,y’,y’’) = y’’ + y’ – 6y = 0
We have g’(x) = 2e
2x
and g’’(x) = 4e
2x
, when we substitute in the equation, we get
F(x, g(x), g’(x), g’’(x))= 4e
2x
+ 2e
2x
– 6 e
2x
= 0 and defined for all x real.
2) Verify that the function h(x) =
x
2
3
defined on the interval (0, 3) is an explicit solution
of the equation F(x,y,y’) = 3xy’ – 2y = 0.
Since g’(x) =
2
3
x
!
1
3
is defined on the interval (0, 3) and
F(x, g(x), g’(x)) =
3x
2
3
x
!
1
3
!
2x
2
3
=
2x
2
3
!
2x
2
3
=
0
and defined for all x in (0, 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