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From: ken@turtlevax.UUCP (Ken Turkowski)
Newsgroups: net.math
Subject: Re: please help with diffeq
Message-ID: <526@turtlevax.UUCP>
Date: Sat, 22-Sep-84 03:15:22 EDT
Article-I.D.: turtleva.526
Posted: Sat Sep 22 03:15:22 1984
Date-Received: Wed, 26-Sep-84 04:12:36 EDT
References: <563@ihlts.UUCP>
Organization: CADLINC, Inc. @ Palo Alto, CA
Lines: 48

> I came across this differential equation recently and, being a few years
> out of practice with differential equations, found I could not immediately
> solve it.  I'll appreciate any suggestions or comments you all have.
> 		f(t) * [f"(t) + C] = K
> where f(t) is a real-valued function of a single real variable, f"(t) is
> the second derivative of this function w.r.t. t and C and K are positive
> real constants.  We also know that f(0) = d > 0 and f'(0) = 0.  Help?
> --
> Roger Noe			ihnp4!ihlts!rjnoe

This second order differential equation can be transformed into a first
order equation as follows:

First, let the parameter be implicit, i.e. f == f(t), let int stand for
the integral, and x^2 stand for x squared.  Then we have:

f [ f" + C ] = K

f" + C = K/f

f" = K/f - C

Let us multiply by f':

f'f" = K f'/f - Cf'

Integrate:

int { f'f" dt } = K int { f'/f dt } - C int { f' dt }

int { f' d(f') } = K int { 1/f df } - C int { df }

(1/2) (f') ^ 2 = K ln(f) - C f + B

Reintroducing the parameter:

-----------------------------------------------------------------

(1/2) [f'(t)] ^ 2 = K ln(f(t)) - C f(t) + B

-----------------------------------------------------------------

Now the problem is to solve the above first order differential equation.

-- 
Ken Turkowski @ CADLINC, Palo Alto, CA
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