[net.math] Differential equation

flinn@seismo.UUCP (E. A. Flinn) (12/02/83)

The following problem appeared in Technology Review recently:
Solve the infinite-order ordinary differential equation

f = f' + 2f" + 3f'" + 4f"" + ...

mkr@CS-Arthur (Mahesh K Rathi) (12/04/83)

The solution to the infinite order differential equation

  f = f'+2f''+3f'''+ ......


  is  f(x) = exp(rx) where  r = (3-sqrt(5))/2, which is easily obtained
  by putting  f=exp(rx) in the diff. eqn.


  How did I guess that the solution should be of the type f=exp(rx) ?
  Well, by noting that the soln. to  f=f'  is exp(x), the soln. to
  f=f'+2f''  is exp(x/2) etc. etc.

wolit@rabbit.UUCP (12/05/83)

If f(x) = 0, then

f = f' + 2f" + 3f'" + 4f"" + ...
  = 0

	Jan Wolitzky, AT&T Bell Labs, Murray Hill, NJ

apdoo@alice.UUCP (12/05/83)

	There is, of course, a symbolic way of solving this equation.
Writing D for the differential operator, we have
  inf   n
f=sum nD f
  n=1

so
    Df
f=---------
   (1-D)**2
Solving,
(1-D)**2 f =Df

f-3Df+D**2 f =0

f=c exp(rx) + k exp(sx)

where c and k are arbitrary constants and r and s are the roots of
1-3x+x**2=0

or (3+-sqrt(5))/2