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Solving a Differential Equation

Date: 6/5/96 at 21:32:37
From: Anonymous
Subject: Elementary Solution?

Dr. Math:

  When I solved the first order differential equation

           x(y')+xy = 1           (*)

 by dividing by x, then multiplying by the integrating factor,
 and obtained a solution which required me to solve an integral
 looking somewhat like this

            S ((e^u)/u) du

 the integral resisted every attack on my part.

 I was able to iteratively apply a modified version of
 integration by parts to the integral above, obtaining the
 following series:

           00  (<--infinity)                    
          \      (k-1)!  
  (e^u) *  .    -------                         (***)
          /       u^k
          k = 1

 Intuitively, I would say the differential equation (*) has no 
solution expressable in terms of elementary functions. How would I go
about finding out about this? In other words, is there a way to 
*prove* that (***) above does not belong to the set of elementary

 I would like to note the resemblance of the summation part in (***) 
to that of the Taylor series expansion of e^z
           \        z^(k-1) 
   e^z =   .       --------   
           /         (k-1)!
           k = 1

 So, is there a manner to solve the diff.eq. (*) such that
an elementary form of (***) will be obtained?

P.S. In case you want to know, the version of the integration
by parts I used is:
      S (1/a) db  =  a/b  +  S (a/b^2) da

 which can be proven quite easily.

Date: 6/6/96 at 6:13:59
From: Doctor Anthony
Subject: Re: Elementary Solution?

There is in fact a solution in series to the differential equation

x(dy/dx) + xy = 1    
   dy/dx + y = 1/x
e^(x)(dy/dx) + e^(x)*y = e^(x)/x   and integrating we get

   y*e^x = INT[e^(x)dx/x]

         = ln|x| + x/1! + x^2/(2.2!) + x^3/(3.3!) + x^4/(4.4!) + ...

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
College Calculus
High School Calculus

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