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 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