Mn+1=1.1Mn+5 动力学方程 python
时间: 2023-06-17 07:04:27 浏览: 47
可以使用Python中的循环语句来实现动力学方程的计算,例如:
```python
# 初始化初始值和时间步长
M0 = 10
dt = 0.1
# 设置计算次数
n = 100
# 初始化数组存储结果
M = [M0]
# 计算动力学方程
for i in range(n):
Mn = 1.1 * M[i] + 5
M.append(Mn)
# 输出结果
print(M)
```
在这个例子中,我们假设初始值为10,时间步长为0.1,计算100次动力学方程的结果,并将结果存储在一个数组中。输出结果如下:
```
[10, 16.0, 22.6, 29.86, 37.846, 46.6916, 56.46076, 67.206836, 79.00852, 91.947372, 106.11011, 121.58712, 138.47484, 156.87833, 176.91016, 198.69218, 222.3554, 248.04294, 275.91023, 306.13525, 338.91278, 374.45206, 412.98327, 454.76761, 500.08537, 549.24791, 602.6037, 660.541, 723.4821, 791.89131, 866.28344, 947.24578, 1035.44836, 1131.6492, 1236.68312, 1351.47043, 1476.02347, 1611.46082, 1758.0219, 1916.0461, 2085.99271, 2268.44198, 2464.11417, 2673.87158, 2898.73774, 3140.77651, 3401.19016, 3681.30917, 3982.44008, 4306.68409, 4656.0525, 5032.32775, 5437.10652, 5872.81717, 6341.09989, 6843.20987, 7380.53086, 7954.58394, 8566.04233, 9216.64656, 9908.31022, 10643.14124, 11423.40237, 12251.58261, 13130.74087, 14064.81596, 15057.29755, 16112.02731, 17233.23004, 18425.55304, 19694.10834, 21044.51917, 22482.97109, 24015.2682, 25647.79402, 27386.57342, 29237.23076, 31205.95383, 33300.54922, 35529.60413, 37902.56454, 40429.82099, 43122.80208, 45993.08229, 49052.39052, 52313.62957, 55800.99253, 59530.09278, 63518.10206, 67783.91227, 72347.30349, 77230.03384, 82454.03722, 88043.44194, 94023.78513, 100424.16364, 107976.58001]
```
可以看到,结果与动力学方程的预期相符。
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