答案是对的,例如
syms x y
>> da=dsolve('Dx=y+1,Dy=x+1','x(0)=-2,y(0)=1')
da =
x: [1x1 sym]
y: [1x1 sym]
这就是你说的情况了。如要详细一点,用指令:
>> da.x
ans =
1/2*exp(t)-3/2*exp(-t)-1
>> da.y
ans =
1/2*exp(t)+3/2*exp(-t)-1
>> [a,b]=dsolve('Dx=y+1','Dy=x+1','x(0)=-2')
a =
-exp(-t)*(C3 + exp(2*t)*(exp(-t) - C3 + 1))
b =
exp(-t)*(C3 - exp(2*t)*(exp(-t) - C3 + 1))