5楼: Originally posted by change0618 at 2013-09-11 11:35:44
这样也可以

function a = myji(m)
myfun = @(r,m)power(power(r,5)-power(m,5),-1./2);
a = quadl(@(r)myfun(r,m),m,973.15);
end
...

2楼: Originally posted by change0618 at 2013-05-20 14:29:33
边值问题，而且是二阶的，先变化成一阶方程组，然后用bvp4c解决

3楼: Originally posted by 转石千仞 at 2013-05-21 14:24:09
M文件：
function =ji(m)
a=quad('power(power(r,5)-power(m,5),-1./2)',m,973.15)
－－－－－－－－－－－－－－－－－－－－－－－－－－－
function =ji(m)
a=quad('power(power(r,5)-power(50,5),-1./2)' ...

function MyInteger
clear all
clc
m = 10:10:50;
for i = 1:length(m)
    a(i) = quadl(@(t)myfun(t,m(i)),m(i),973.15);
end
a
end
function f = myfun(t,m)
f = power(power(t,5)-power(m,5),-1./2);
end