变系数非线性常微分方程组用mathematica怎么求解?
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/22 15:55:13
变系数非线性常微分方程组用mathematica怎么求解?
![变系数非线性常微分方程组用mathematica怎么求解?](/uploads/image/z/16802076-12-6.jpg?t=%E5%8F%98%E7%B3%BB%E6%95%B0%E9%9D%9E%E7%BA%BF%E6%80%A7%E5%B8%B8%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E7%BB%84%E7%94%A8mathematica%E6%80%8E%E4%B9%88%E6%B1%82%E8%A7%A3%3F)
不要理会那些广告商.
用DSolve函数.
如输入:
DSolve[y'[x] + x y'[x]^2 == 1,y,x]
输出:
{{y -> Function[{x},
C[1] + 1/2 (-2 Sqrt[1 + 4 x] - 2 Log[-1 + Sqrt[1 + 4 x]])]},{y ->
Function[{x},
C[1] + 1/2 (2 Sqrt[1 + 4 x] - 2 Log[1 + Sqrt[1 + 4 x]])]}}
再如输入:
DSolve[ {y'[x] == Exp[z[x]] + 1,z'[x] == y[x] - x},{y,z},
x] // Quiet
输出:
{{z -> Function[{x},
Log[C[1] +
C[1] Tan[
1/2 (Sqrt[2] x Sqrt[C[1]] + 2 Sqrt[2] Sqrt[C[1]] C[2])]^2]],
y -> Function[{x},
x + Sqrt[2] Sqrt[C[1]]
Tan[1/2 (Sqrt[2] x Sqrt[C[1]] + 2 Sqrt[2] Sqrt[C[1]] C[2])]]}}
用DSolve函数.
如输入:
DSolve[y'[x] + x y'[x]^2 == 1,y,x]
输出:
{{y -> Function[{x},
C[1] + 1/2 (-2 Sqrt[1 + 4 x] - 2 Log[-1 + Sqrt[1 + 4 x]])]},{y ->
Function[{x},
C[1] + 1/2 (2 Sqrt[1 + 4 x] - 2 Log[1 + Sqrt[1 + 4 x]])]}}
再如输入:
DSolve[ {y'[x] == Exp[z[x]] + 1,z'[x] == y[x] - x},{y,z},
x] // Quiet
输出:
{{z -> Function[{x},
Log[C[1] +
C[1] Tan[
1/2 (Sqrt[2] x Sqrt[C[1]] + 2 Sqrt[2] Sqrt[C[1]] C[2])]^2]],
y -> Function[{x},
x + Sqrt[2] Sqrt[C[1]]
Tan[1/2 (Sqrt[2] x Sqrt[C[1]] + 2 Sqrt[2] Sqrt[C[1]] C[2])]]}}