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Re: NDSolve[]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg123536] Re: NDSolve[]
  • From: Murat Havzalı <gezginorman at gmail.com>
  • Date: Sat, 10 Dec 2011 07:30:21 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201112091058.FAA03928@smc.vnet.net>

I think that is somewhat about the usage of the built-in function
"derivative". What I understand from the "help" file, you have to define the
function you want to take derivative of first, if you want to use the
built-in function "derivative".

The equivalent of the second one is as follows:

NDSolve[{D[x[t], {t, 1}] == -y[t] - x[t]^2, 
  D[y[t], {t, 1}] == 2*x[t] - y[t], x[0] == y[0] == 1}, {x, y}, {t, 
  10}]

Which, supposedly gives the same answer as 

NDSolve[{x'[t] == -y[t] - x[t]^2, y'[t] == 2*x[t] - y[t], 
  x[0] == y[0] == 1}, {x, y}, {t, 10}]

Best,

H

-----Original Message-----
Sent: Friday, December 09, 2011 12:58 PM
Subject: [mg123536] NDSolve[]
To: mathgroup at smc.vnet.net

For the same equations, why does the first method as following give the
error but the other one give the desired result?

NDSolve[{Derivative[x[t], t] == -y[t] - x[t]^2,
  Derivative[y[t], t] == 2*x[t] - y[t], x[0] == y[0] == 1}, {x,
  y}, {t, 10}]

NDSolve[{x'[t] == -y[t] - x[t]^2, y'[t] == 2*x[t] - y[t],
  x[0] == y[0] == 1}, {x, y}, {t, 10}]

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