From: wesson@egr.msu.edu (Dale Wesson)
Newsgroups: sci.math
Subject: Closed form solution of 1st order ODE.
Date: Mon, 25 Mar 96 01:29:30 GMT

Hi there,

I was wondering whether anybody knows of a closed form solution
to the following first order ODE:

 dy/dx = a/y + b y/x + c/(x)^0.5

a,b and c are multiplicative constants, nothing to worry about.
If a=0 above eq. reduces to y'+Py+Q=0, for which the solution is
well known. For c=0 above eq. resembles Bernoulli's eq.
But with the one as above, I am stuck.

I'd appreciate any comments about possible ways to obtain a closed
form solution. My E-Mail is: weispfen@egr.msu.edu

Thanks a lot in advance.

Klaus Weispfennig

==============================================================================

From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Closed form solution of 1st order ODE.
Date: 28 Mar 1996 06:00:05 GMT

In article <4j4t7l$pa4@msunews.cl.msu.edu>,
Dale Wesson <wesson@egr.msu.edu> wrote:

>I was wondering whether anybody knows of a closed form solution
>to the following first order ODE:
>
> dy/dx = a/y + b y/x + c/(x)^0.5

If  y  is a solution then the function  v = (x)^0.5 / y  satisfies
(after a bit of simplification)

v' = (1/x)( a v^3 + c v^2 + (b+1/2) v),

a separable differential equation. If the roots of this polynomial in  v
are  0,  r, and  s, then we deduce

ln|v| / (b+1/2) + (1/a)(ln|v-r|/r(r-s) + ln|v-s|/s(s-r)) = ln|x| + C 

for some constant  C, so that

|v|^(1/(b+1/2)) * |v-r|^(1/(ar(r-s))) * |v-s|^(1/as(s-r)) = A x

for some constant  A. Those three exponents sum to zero, so this is
equally well written

|y-sqrt(x)/r|^(1/(ar(r-s))) * |y-sqrt(x)/s|^(1/as(s-r)) = A' x^(2b/(2b+1))

for some constant  A'. The final form of the solution is a matter of
taste, but I don't think that in general you can write it with any nicer
reference to  y.

(Obviously if the roots  r  and  s  are complex, you might prefer the
arctan form of the integral used to solve the separable equation.)

dave