谁能看看下边的程序怎么回事？

lvhaiwei007

S = dsolve('Df = f + g','Dg = -f + g','f(0) = 1','g(0) = 2')

S =

    f: [1x1 sym]
    g: [1x1 sym]

正确结果为
S.f = (i + 1/2)/exp(t*(i - 1)) - exp(t*(i + 1))*(i - 1/2)
         S.g = exp(t*(i + 1))*(i/2 + 1) - (i/2 - 1)/exp(t*(i - 1))
为什么出错了呢？

bainhome

你又是通过什么判断出MATLAB结果出错了？

1. >> S = dsolve('Df = f + g','Dg = -f + g','f(0) = 1','g(0) = 2')
   
2. S =
   
3.     f: [1x1 sym]
   
4.     g: [1x1 sym]
   
5. >> S.f
   
6. ans =
   
7. exp(t)*cos(t) + 2*exp(t)*sin(t)
   
8. >> S.g
   
9. ans =
   
10. 2*exp(t)*cos(t) - exp(t)*sin(t)
    
11. >> diff(S.f)
    
12. ans =
    
13. 3*exp(t)*cos(t) + exp(t)*sin(t)
    
14. >> S.f+S.g
    
15. ans =
    
16. 3*exp(t)*cos(t) + exp(t)*sin(t)
    
17. >> syms t
    
18. >> subs(S.f,t,0)
    
19. ans =
    
20.      1
    
21. >> subs(S.g,t,0)
    
22. ans =
    
23.      2
    
24. >> -S.f+S.g
    
25. ans =
    
26. exp(t)*cos(t) - 3*exp(t)*sin(t)
    
27. >> S.g
    
28. ans =
    
29. 2*exp(t)*cos(t) - exp(t)*sin(t)
    
30. >> diff(S.g)
    
31. ans =
    
32. exp(t)*cos(t) - 3*exp(t)*sin(t)

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It looks perfect to me.

tenglang

结果和楼上的只是表达形式不同. 一个是用三角函数表示,一个是用复数表示.
help dslove, you will find the exmaples in the help

1. Examples:
   
2. 
   
3. dsolve('Dx = -a*x') returns
   
4. 
   
5. ans = C1/exp(a*t)
   
6. 
   
7. x = dsolve('Dx = -a*x','x(0) = 1','s') returns
   
8. 
   
9. x = 1/exp(a*s)
   
10. 
    
11. S = dsolve('Df = f + g','Dg = -f + g','f(0) = 1','g(0) = 2')
    
12. returns a structure S with fields
    
13. 
    
14. S.f = (i + 1/2)/exp(t*(i - 1)) - exp(t*(i + 1))*(i - 1/2)
    
15. S.g = exp(t*(i + 1))*(i/2 + 1) - (i/2 - 1)/exp(t*(i - 1))

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lvhaiwei007

回复 3 # tenglang 的帖子

谢谢，对Matlab不熟