# Ecuaciones diferenciales con polymath

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• Publicado : 16 de junio de 2011

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1.

POLYMATH Results

ODE Report (RKF45)

Differential equations as entered by the user
[1] d(y)/d(t) = z
[2] d(z)/d(t) = t+y

Explicit equations as entered by the userIndependent variable
variable name : t
initial value : 0
x | y | y calculada |
2.0861613 | -1 | -1 |
3.2613625 | 0 | 0 |
2.5 | | -0.64785715 |
final value : 1

Calculatedvalues of the DEQ variables

Variable initial value minimal value maximal value final value
t 0 0 1 1
y 2 23.2613625 3.2613625
z 0 0 2.893483 2.893483
Variable initial value minimal value maximal value final value
t 00 1 1
y 2 1.754857 2.0861613 2.0861613
z -1 -1 1.3504024 1.3504024
Variable initialvalue minimal value maximal value final value
t 0 0 1 1
y 2 1.8977974 2.5 2.5
z-0.6478572 -0.6478572 1.8937872 1.8937872

* y(1)= 2.5
z(0)= -0.64785715

2.
POLYMATH Results

Differential equations as entered by the user
[1] d(y)/d(t) = z
[2]d(z)/d(t) = -2*t*z

Explicit equations as entered by the user

Independent variable
variable name : t
initial value : 0
final value : 10

x | y | y calculada |
0.1137731 | -1 | -1 |1.8862269 | 1 | 1 |
1 | | 0 |

Variable initial value minimal value maximal value final value
t 0 0 10 10
y 10.1137731 1 0.1137731
z -1 -1 -1.652E-44 -1.652E-44

Variable initial value minimal value maximal value final value
t 0...