Ordinary differential
equations (ODE’s)
definitions of ODE, initial (boundary) conditions,
general and particular solutions of an ODE
integration of some 1st order diff. equations
integration of some 2nd order ODE’s
diff. eq. for a harmonic oscillator
1st integral and its relation to the energy
conservation law in mechanics
Literature:
All you wanted to know about Mathematics,
but were afraid to ask, L.Lyons - Volume 1, Chapt. 5
Other reading:
Mathematical Methods for Science Students, G.Stephenson – Chapt. 21
Mathematical Methods in Physical Sciences, M.L.Boas – Chapt. 8
An ordinary differential equation (ODE),
 d n x d n1 x

dx
 n , n1 ,  , , x, t   0
dt
 dt dt

relates values of derivatives of an unknown function x(t) of variable t,
dx d 2 x
d nx
, 2 ,  , n
dt dt
dt
x(t ), x' (t ),  , x( n ) (t )
to values of this function x(t) for each value of variable t.
The order n of the highest derivative present in such an equation
determines the order of a differential equation.
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Example:
Radioactive decay is described by the 1st order ordinary differential equation:
dx
x

dt

x(t) – remaining density of radioactive nuclei
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Example: Newton’s equation is the 2nd order ordinary differential equation:
d 2x
 dx

m 2  F  , x, t 
dt
 dt

dv(t ) d 2 x(t )
Reminder: acceleration 
dt
dt 2
x(t) – coordinate of a particle
velocity - v (t ) 
dx (t )
dt
Methods of solving 1st order ODE’s
General form of the 1st order ODE:
where F[x,t] can include any
dependence on x and t.
dx
 F [ x(t ), t ]
dt
dx(t )
t
dt
2. Direct integration of ODE’s:
one applies the operation of integration
t
 dt  to both sides:
0
t
t
t2
0 tdt  2
dx
0 dt dt  x(t )  x(0)
solution:
t2
x(t )  x(0) 
2

initial
condition
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more general:
dx (t )
 f (t )
dt

t
x(t )  x(0)   dt f (t )
0
For revision, read in FLAP, modules PHYS 114 and 115
Relation
between operations of integration and differentiation
to remember:
t
for any function
x(t),
dx
0 dt dt  x(t )  x(0)
Integration and differentiation are inverse to each other.
More general:
for any function G(x)
d

dt
G
[
x
(
t
)]
0  dt
  G[ x(t )]  G[ x(0)]

t
2. ODE’s with 'separable variables':
Example:
dx
 ax
dt
1 dx
 a  const
x dt
d
1 dx
ln[ x(t )] 
dt
x dt
d
ln[ x(t )]  a
dt
t
As before, one applies the operation of integration  dt  to both sides:
0
t
t
d
x(t )
dt
ln[
x
(
t
)]

ln[
x
(
t
)]

ln[
x
(
0
)]

ln
0 dt
x(0)
 adt  at

x(t )  x(0)  e at
x(t )
ln
 at
x ( 0)
exponentiate
both sides
0
initial
value
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*)
We have proven systematically that the 1st order linear ODE
dx
 ax  0
dt
linear
has, indeed, a general solution
x(t )  A  eat
determined by
initial condition
General form of a separable 1st order ODE
dx f (t )

dt g ( x)
integrate t
dt 
both sides: 0
for any pair of functions
f(t) and g(x)
dx
g[ x(t )]  f (t )
dt
t
dx
0 dt dt g[ x(t )] 
dt 
change of variables
of integration in the
integral over t on
the left hand side
dG
 g (x)
dx
t
 dt
f (t )
0
dx(t )
 dx(t )
dt
x (t )
t
 dx g ( x)   dt
x (0)
f (t )
0
x
G( x)  G( x(0)) 
 dx g ( x)
x ( 0)
This reduces the differential equation problem to the solution of an algebraic
equation (non-differential), which would relate the value of function x(t) for
each t to the value of t and one free parameter x(0) (initial condition).
x (t )
t
 dx g ( x)  G[ x(t )]  G[ x(0)]   dt
x (0)
0
f (t )
Example:
dx
  x 2et
dt

 f (t )  e t


1
g
(
x
)



x2

f (t )
g ( x)
x (t )
t
 dx g ( x)   dt
x (0)
x (t )
f (t )
0
x (t )
dx
1
1
dx
g
(
x
)



x(0)
x(0)  x 2 x(t ) x(0)
t
t
0
0
t
t
dt
f
(
t
)

dt
e

e
1


1
1

 et  1
x(t ) x(0)
1
x(t ) 
et 1 
1
x ( 0)
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Quick check:
x ( 0) 
1
1 1 
1
x(0)

1
 x(0)
1 / x(0)
3.
Home reading
‘Scale-invariant’ equations:
equations which do not change their form
upon a simultaneous re-scaling of both x and t
can be reduced to separable ODE’s
dx
 F [ x(t ), t ]
dt
t  at

 x  ax
F [ax, at ]  F [ x, t ]
with
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Example:
dx xt  x 2

dt
t2
(note that this equation is not separable)
Let us introduce a new intermediate unknown function
x(t )
u (t ) 
t

x(t )  tu(t )
dx d (tu(t ))
du

 u t
dt
dt
dt
xt  x 2 ut 2  ut 
2



u
u
2
2
t
t
2
du
2
u

t


u

u
 dt

du
u2

dt
t
separable equation
du
1 tt
dt
  u 2  u  xx  lnln tt  AA   t
General solution:
t
x(t ) 
ln t  A  parameterised by one 'free' parameter
Initial (boundary) conditions in ODE’s
General solution
1sr order ODE

dx
 ax
dt
x(t )  A  e
at
General solution of the 1st order ODE contains one 'free' parameter.
It describes a group of all possible functions, which obey some given ODE.
When solving a particular physical problem, one seeks a particular solution
related to given initial (boundary) conditions. This requires specifying the
value of a 'free' parameter, thus choosing one function from many: the one
obeying additional conditions on the beginning of a studied interval of
variable t.
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Example:
 dx
  5x
 dt
 x(0)  2

5t general
 x(t )  Ae

 x(0)  2
solution

A  e0  A  2
Particular solution, which satisfies the initial condition: x(t )  2e
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5t
Differential equation
without initial (boundary)
conditions

General solution
Differential equation
with initial (boundary)
conditions

A particular solution
fully defined
dynamical
problem

differential
equation

initial
(boundary)
conditions
To determine only one definite particular solution,
any ordinary differential equation of the n-th order,
d x d x

dx
 n , n1 ,  , , x, t   0
dt
 dt dt

n
n 1
has to be complemented with n equations which
relate the values of x(t) and
dx d 2 x
d n1 x its lower-order derivatives,
x, , 2 ,  , n1 taken at the ends of a
dt dt
dt
studied interval, a<t<b.
In practice, one has to obtain the general solution
xa1a2 an (t )
of a differential equation, first. The latter would depend
on n free parameters. To find a particular solution,
one identifies the values of all these n parameters,
1
2
n using boundary conditions.
a , a ,  , a