查看官方代码有类似的例子ifun = First[u /. NDSolve[{u''[t] + u[t] == 0, u[0] == 0, u'[0] == 1}, u, {t, 0, \[Pi]}]]，有点看不懂这个First函数的意思。我的是一个微分方程组，我该如何随意调用数值解的函数，得到数值解函数任一点值和导数值呢。谢谢

U = 0.3 340; b = 0.5; k1 = 0.1; \[Theta] =
15 Pi/180 + 10 Pi/180 Sin[k t]; a = -0.1; \[Rho] = 1.225; cc =
2 b; k = k1 U/b;(*k1为减缩频率*)
NN = 8; Di =
Table[Which[i == j + 1, 1/2/i, i == j - 1, -1/2/i, True, 0], {i,
NN}, {j, NN}];
bn = Table[((-1)^(i - 1)*((NN + i - 1)!))/((NN - i - 1)!*(i!)^2), {i,
NN - 1}];
bn = AppendTo[bn, -1];
d = Prepend[Table[0, {i, NN - 1}], 1/2];
c = Table[2/i, {i, NN}];
A = Di + Transpose[{d}].{bn} + Transpose[{c}].{d} +
1/2 Transpose[{c}].{bn};
\[Lambda] = {\[Lambda]1[t], \[Lambda]2[t], \[Lambda]3[t], \[Lambda]4[
t], \[Lambda]5[t], \[Lambda]6[t], \[Lambda]7[t], \[Lambda]8[t]};
equ1 = A.D[\[Lambda], t] + U/b Transpose[{\[Lambda]}] ==
Transpose[{c}] (U D[\[Theta], t] + b (1/2 - a) D[\[Theta], {t, 2}]);
ans = NDSolve[{equ1, {\[Lambda]1[0], \[Lambda]2[0], \[Lambda]3[
0], \[Lambda]4[0], \[Lambda]5[0], \[Lambda]6[0], \[Lambda]7[
0], \[Lambda]8[0]} == {2.6516624792177117`, 1.4392866441021448`,
0.9672678190313838`, 0.7267703507900245`, 0.58
18098195675
75`,
0.48500273616324474`, 0.4
15810058570
8028`,
0.3633991127816457`}}, {\[Lambda]1[t], \[Lambda]2[t], \[Lambda]3[
t], \[Lambda]4[t], \[Lambda]5[t], \[Lambda]6[t], \[Lambda]7[
t], \[Lambda]8[t]}, {t, 0, 2 Pi/k}]
{\[Lambda]1[t], \[Lambda]2[t], \[Lambda]3[t], \[Lambda]4[
t], \[Lambda]5[t], \[Lambda]6[t], \[Lambda]7[t], \[Lambda]8[
t]} /. ans /. t -> 2 Pi/k (*寻找初值*)
(*Plot[Evaluate[{\[Lambda]1[t],\[Lambda]2[t],\[Lambda]3[t],\[Lambda]4[\
t],\[Lambda]5[t],\[Lambda]6[t]}/.ans],{t,0,2 Pi/k}]*)
\[Lambda]0 = 1/2 bn.\[Lambda];
L0 = Pi \[Rho] b^2 (U D[\[Theta], t] - b a D[\[Theta], {t, 2}]) +
2 Pi \[Rho] U b (U \[Theta] +
b (1/2 - a) D[\[Theta], t] - \[Lambda]0);
\[Alpha] = \[Theta] + b/U (1/2 - a) D[\[Theta], t] - \[Lambda]0/U;
方程组的解为Lamda[1]至Lamda[8],我最终是要得到任一时间点的Alpha，和Alpha'

……可参看Stackexchange帖子《Derivative of function solved from NDSolve can't be plotted》，链接就不贴了，省得被吞。

好的，十分感谢。百度竟然没给我推送通知，刚看到