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【答案】应助回帖
★ ★ ★ ★ ★ ★ ★ 感谢参与,应助指数 +1 caemechanics: 仿真EPI+1, 感谢应助 2012-03-28 21:48:58 bdwhq: 金币+7, ★★★★★最佳答案, 非常感谢dbb627 的热心帮助,谢谢谢谢!!!!!!! 2012-03-29 08:32:26
>> x1=sym('x1(t)');x2=sym('x2(t)');
>> vv=-5*(x2+10*x1+x1*sin(x1))-x1*x2
vv =
- 50*x1(t) - 5*x2(t) - x1(t)*x2(t) - 5*sin(x1(t))*x1(t)
>> syms t
>> dv=diff(vv,t)
dv =
- x1(t)*diff(x2(t), t) - x2(t)*diff(x1(t), t) - 5*sin(x1(t))*diff(x1(t), t) - 50*diff(x1(t), t) - 5*diff(x2(t), t) - 5*cos(x1(t))*x1(t)*diff(x1(t), t)
================================
v‘=-x1*x2'-x2*x1'-5*sin(x1)*x1'-50*x1'-5*x2'-5*cos(x1)*x1*x1'
微分方程组变为
x1=x(1)
x2=x(2)
v=x(3)
w=x(4)
===========
x1’=x2+x1*sin(x1);
x2‘=w+x1*x2;
v‘=-x1*x2'-x2*x1'-5*sin(x1)*x1'-50*x1'-5*x2'-5*cos(x1)*x1*x1'
w'=|v'|*(0.5*v-w)+0.3*v';
matlab 代码
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