MATH2310 – DE Lecture 4
Non-solvable DE’s:
Consider the DE:
( y ' )2 +1=0where y ( x ) is a real-valued function.
This DE does not have a solution.
Solutions of IVP’s vs DE’s:
Consider the DE:
'
y =y
There are many solutions, however each IVP with y ( 0 )= y 0 has exactly one solution.
This does not hold true for all DE’s however. For example, the DE:
dx
=√ x , x ( 0 ) =0has multiple solutions.
dt
Equilibrium solutions:
Given
dx
=f (x , t)Then all x for all t ϵ R , f ( t , x ) =0 are equilibrium points of the ODE.
dt
For example, the DE:
dx
=x( x−1)has equilibrium points x=0 , 1.
dt
The functions x 1 ( t )=0 , x 2 ( t )=1 are both constant solutions of the DE.
Existence and uniqueness theorem:
∂f
If f and are continuous on some rectangle in the (x , y )-plane, centred on (x 0 , y 0 ),
∂y
then the IVP
dy
=f ( x , y ) , y ( x0 ) = y 0has a unique solution which is continuously differentiable on at
dx
least |x−x 0|< h, where h> 0.
Non-solvable DE’s:
Consider the DE:
( y ' )2 +1=0where y ( x ) is a real-valued function.
This DE does not have a solution.
Solutions of IVP’s vs DE’s:
Consider the DE:
'
y =y
There are many solutions, however each IVP with y ( 0 )= y 0 has exactly one solution.
This does not hold true for all DE’s however. For example, the DE:
dx
=√ x , x ( 0 ) =0has multiple solutions.
dt
Equilibrium solutions:
Given
dx
=f (x , t)Then all x for all t ϵ R , f ( t , x ) =0 are equilibrium points of the ODE.
dt
For example, the DE:
dx
=x( x−1)has equilibrium points x=0 , 1.
dt
The functions x 1 ( t )=0 , x 2 ( t )=1 are both constant solutions of the DE.
Existence and uniqueness theorem:
∂f
If f and are continuous on some rectangle in the (x , y )-plane, centred on (x 0 , y 0 ),
∂y
then the IVP
dy
=f ( x , y ) , y ( x0 ) = y 0has a unique solution which is continuously differentiable on at
dx
least |x−x 0|< h, where h> 0.
