rjnoe@ihlts.UUCP (Roger Noe) (09/19/84)
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
ken@turtlevax.UUCP (Ken Turkowski) (09/22/84)
> 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 UUCP: {amd,decwrl,dual,flairvax,nsc}!turtlevax!ken ARPA: turtlevax!ken@DECWRL.ARPA