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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- References:
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- From: "Steven M. Christensen" <steve@smc.vnet.net>
- [no subject]