A General Method Of Finding A Particular Solution To The
Second Order Linear Ordinary Differential Equation
                           By Ronald H. Brady

The presentation that follows is an informal preliminary draft.

In what follows all of the integrals involved will be (iterated) indefinite
integrals of  functions of a single variable. All constants of integration
will be set equal to zero."Dummy" variables of integration will not be
used. Instead we write, when  S = S(x) is an integrable function of x,

∫∫Sdx^2  <=>   ∫(∫Sdx)dx                                                                       (eq.1a)

∫∫S ∫∫Sdx^4  <=> ∫(∫(S (∫∫Sdx^2))dx)dx                                                  (eq.1b)

etc.

Please note that we have used <=> to denote the identity symbol.

Let  y = y(x) be function of the variable x. Also let
y' and y'' denote the first and second derivatives, of y with respect
to x, respectively. Now let

y''  +  Sy   =  0                                                                                      (eq.2)

It is proposed that a particular solution of (eq.2) is given by

y = 1 - ∫∫Sdx^2 +  ∫∫S ∫∫Sdx^4 -  ∫∫S∫∫S ∫∫Sdx^6  + ....                                                                                                                                                    (eq.3)
provided that the indicated integrals are defined and the
infinite series (on the right side of (eq.3)) converges.

The proof  is a straight forward exercise and will be
given later.

Let's illustrate by a specific example. Given the equation

y''   +  (x^k)y = 0                                                                                  (eq.4)

where k is a constant.
In this particular case we have S = x^k .  Substituting for S
in eq.(3) and performing  the indicated operations will bring us to

y = 1 - x^(k+2)/[(k+1)(k+2)] + x^(2k+4))/[(k+1)(k+2)(2k+3)(2k+4)]

       - x^(3k+6)/[(k+1)(k+2)(2k+3)(2k+4)(3k+5)(3k+6)] + ...             (eq.5)

We will denote the first term on the right of  (eq.5) by u_0 .
Therefore

u_0  =  1

We will denote general or the nth term by u_n  
(for n greater > or = to 1) It will be seen that the nth term may
be given by

u_n = x^(n(k+2))/[(k+1)(k+2)(2k+3)(2k+4)...(n(k+2)-1)(n(k+2)] 
                                                                                                           (eq.6)
To remove any possible ambiguity we may also write u_n as
follows:

u_n =  x^(n(k+2))/[P_i{(i(k+2)- 1)(i(k+2))}]                                      (eq.6a)

where the product operator P_i{ .....} indicates the product of
the quantity in the curly braces: with the index i running from i =1
to i =n .

There are two "general factors" in the denominator on the right side of (eq.6)  
(n(k+2) - 1) and (n(k+2)). To insure that the denominator does not vanish
we write

n(k+2) - 1  "not =" 0  for n=1,2,3,.                       

where the symbol "not =" is used in lieu of the standard symbol
for "not equal to"                                                        

Solving this inequality for k will result in

k "not = " (1/n)  -  2      for n = 1, 2, 3,....                                            (ineq.7)
 
We also write, for the other general factor,

n(k+2) "not =" 0                                                                                                        
from which we have

k "not =" - 2                                                                                         (ineq.8)

Note that the inequality ( 8 ) can be obtained from inequality (7 )
by taking the limit as n approaching infinity.

The convergence of the function defined by (eq.5), for integers k that satisfies
the inequalities,

k "not ="  (1/n) - 2               n =1, 2, 3,...

k "not = " -2

can verified, in a straight forward manner, by using the ratio test to show
that the limit of the ratio of the (n+1)th term to the nth term vanishes in the
limit as n approaches  infinity.

Perhaps we should not be too surprised at the convergence properties of
this series since the exponential function e^x can be written as an
infinite series of integrals

e^x =  1 +   ∫dx  +  ∫ ∫dx^2  +   ∫ ∫ ∫dx^3 +....

e^x =  1 +    x    +   ∫(x)dx      +   ∫ ∫(x)dx^2  +....

e^x  = 1  +   x    +   (1/2)*x^2  +   ∫(1/2)*x^2dx +....


e^x   = 1  +  x     +   (1/2)*x^2  + (1/6)*x^3  + ....  


Now lets consider a very familiar 2nd order differential equation

y''   +    y  = 0                                                                                 (eq.9)

Equa.(4) will reduce to (eq.9) if we set k = 0.

Now setting k = 0 on the right of (eq.5) will result in

y = 1 -  (1/2!)x^2 +  (1/4!)x^4  -  (1/6!)x^6 + ....                           (eq.10)

The reader will recognize the infinite series on the right
as the cosine function which of course is a particular
solution of (eq.9).  

It can therefore be stated that the function defined by (eq.5)
contains the familiar (and very important) cosine function as
a special case.

A Break For Humor(?)

Nothing we must not forget
is the content of the empty set
but the difference between zero and one
is only one more than none.

Now let us consider an nth order O.D.E. of the following
form:

d^(n)y/dx^n    +    Py  =  0                                                             (eq.11)

where the symbol d^(n)y/dx^n denotes the nth derivative
of y with respect to x  and P = P(x) is an integrable function
of x.

In the absence of an equation editor we will use the following
notation:

    ∫ (n)Pdx^n  < = >   ∫ ∫ ... n (times) ∫Pdx^n

to indicate that P is to be "integrated n times".

It is proposed that a particular solution of (eq.11)
is given by

y =  1 -  ∫ (n)Pdx^n  +  ∫ (n)[P∫ (n)Pdx^n]dx^n

          -  ∫ (n)[P∫ (n)[P∫ (n)Pdx^n]dx^n]dx^n + ...                         (eq.12)

provided that the indicated integrals are defined and the
infinite series (on the right side of (eq.12)) converges.

The proof  is a straight forward procedure and will be
presented  later.

Let us conclude with an informal version of the

Theorem Of The Other

If there is one
and if there is another
then surely one of them
must be the other.