我对 odeint 有点困惑.
我在下面找到了一个例子来解决 y"=ay + by'.所以看起来 y[0] 是函数，y[1] 是一阶导数.
I found one example below to solve y"=ay + by'. So it seems that y[0] is the function, y[1] is the first derivative.
下面的表达式是否意味着 y[1] =y' 和 y'[1]= a*y[0]+b*y[1]?
So does the following expression mean y[1] =y' and y'[1]= a*y[0]+b*y[1] ?
如果是y[2], a*y[0]+b*y[1]，它是什么意思?
If it were y[2], a*y[0]+b*y[1], what would it mean?
我有点困惑，因为表达式没有说等式的左侧.
I am a bit confused since the expression does not say the left hand side of the equation.
我也遇到过像[a(y[0], y[1]), b(y[0], y[1])]之类的表达式，但对微分方程一无所知.
I also encountered expressions like [a(y[0], y[1]), b(y[0], y[1])] but have no clue of the differential equation.
这是一个例子:
from scipy.integrate import odeint from pylab import * # for plotting commands def deriv(y,t): # return derivatives of the array y a = -2.0 b = -0.1 return array([ y[1], a*y[0]+b*y[1] ]) time = linspace(0.0,10.0,1000) yinit = array([0.0005,0.2]) # initial values y = odeint(deriv,yinit,time) figure() plot(time,y[:,0]) xlabel('t') ylabel('y') show() 推荐答案让我们在 deriv 中使用 Y 而不是 y 用于其余部分答案要明确:
Let's use Y in deriv instead of y for the rest of answer to be clear:
def deriv(Y,t): # return derivatives of the array Y a = -2.0 b = -0.1 return array([ Y[1], a*Y[0]+b*Y[1] ])函数deriv 以Y = [y, y'] 作为输入.
它应该输出它们的导数([y', y'']).
And it should output their derivatives ([y', y'']).
y' = Y[1]
y'' = a*Y[0]+b*Y[1]