First order diff equation : x' lnx'=1/2 * c

uberathlete

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Jan 16, 2006
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Hi all. I'm really having problems figuring out just how to integrate this equation:

x'(t) * ln[ x'(t) ] = 1/2 * c where c is just some constant.

I thought of using an an integrating factor but the equation doesn't seem to fit the form needed to do so. If anyone can help me out I would really really appreciate it. Thanks!
 
d/dt[x'(t) * ln[ x'(t) ] = 1/2 * c ]

x'(t)[x"(t)/x'(t)] + x"(t)*ln[x'(t)] = 0

x"(t) + x"(t)*ln[x'(t)] = 0

x"(t)[1 + ln[x'(t)]] = 0

x"(t) = 0 ... x'(t) = 1/e

x(t) = (1/e)t + C

back to your original equation ...

x'(t) * ln[ x'(t) ] = 1/2 * c

(1/e)*(-1) = (1/2) * c

c = -2/e
 
skeeter said:
d/dt[x'(t) * ln[ x'(t) ] = 1/2 * c ]

x'(t)[x"(t)/x'(t)] + x"(t)*ln[x'(t)] = 0

x"(t) + x"(t)*ln[x'(t)] = 0

x"(t)[1 + ln[x'(t)]] = 0

x"(t) = 0 ... x'(t) = 1/e

x(t) = (1/e)t + C

back to your original equation ...

x'(t) * ln[ x'(t) ] = 1/2 * c

(1/e)*(-1) = (1/2) * c

c = -2/e

Hmm ... lemme go over this. Thanks skeeter. Much appreciated.
 
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