Matlab求解2元一次方程组syms csyms dti=300R=3.*10.^5k=1.2s=287.*10.^(-6)q0=10h=5qi=0qp1=0qp2=0[c,d]=solve('c=-(q0-qp1-h*(d-ti))/k','-k*(-qi*l/k+c)=h*(-(qi/(2*k))*l^2+c*l+d-ti)-qp2')我这边显示的是The following error occurred converting f
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![Matlab求解2元一次方程组syms csyms dti=300R=3.*10.^5k=1.2s=287.*10.^(-6)q0=10h=5qi=0qp1=0qp2=0[c,d]=solve('c=-(q0-qp1-h*(d-ti))/k','-k*(-qi*l/k+c)=h*(-(qi/(2*k))*l^2+c*l+d-ti)-qp2')我这边显示的是The following error occurred converting f](/uploads/image/z/5171243-59-3.jpg?t=Matlab%E6%B1%82%E8%A7%A32%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%BB%84syms+csyms+dti%3D300R%3D3.%2A10.%5E5k%3D1.2s%3D287.%2A10.%5E%28-6%29q0%3D10h%3D5qi%3D0qp1%3D0qp2%3D0%5Bc%2Cd%5D%3Dsolve%28%27c%3D-%28q0-qp1-h%2A%28d-ti%29%29%2Fk%27%2C%27-k%2A%28-qi%2Al%2Fk%2Bc%29%3Dh%2A%28-%28qi%2F%282%2Ak%29%29%2Al%5E2%2Bc%2Al%2Bd-ti%29-qp2%27%29%E6%88%91%E8%BF%99%E8%BE%B9%E6%98%BE%E7%A4%BA%E7%9A%84%E6%98%AFThe+following+error+occurred+converting+f)
Matlab求解2元一次方程组syms csyms dti=300R=3.*10.^5k=1.2s=287.*10.^(-6)q0=10h=5qi=0qp1=0qp2=0[c,d]=solve('c=-(q0-qp1-h*(d-ti))/k','-k*(-qi*l/k+c)=h*(-(qi/(2*k))*l^2+c*l+d-ti)-qp2')我这边显示的是The following error occurred converting f
Matlab求解2元一次方程组
syms c
syms d
ti=300
R=3.*10.^5
k=1.2
s=287.*10.^(-6)
q0=10
h=5
qi=0
qp1=0
qp2=0
[c,d]=solve('c=-(q0-qp1-h*(d-ti))/k','-k*(-qi*l/k+c)=h*(-(qi/(2*k))*l^2+c*l+d-ti)-qp2')
我这边显示的是
The following error occurred converting from sym to double:
Error using ==> sym.double at 25
DOUBLE cannot convert the input expression into a double array.
If the input expression contains a symbolic variable,use the VPA function instead.
请问是什么问题呢?
1楼,我前面已经给这些赋过值了啊,为什么他不会把这些值带进去呢?
Matlab求解2元一次方程组syms csyms dti=300R=3.*10.^5k=1.2s=287.*10.^(-6)q0=10h=5qi=0qp1=0qp2=0[c,d]=solve('c=-(q0-qp1-h*(d-ti))/k','-k*(-qi*l/k+c)=h*(-(qi/(2*k))*l^2+c*l+d-ti)-qp2')我这边显示的是The following error occurred converting f
syms c
syms d
ti=300;
R=3.*10.^5;
k=1.2;
s=287.*10.^(-6);
q0=10;
h=5;
qi=0;
qp1=0;
qp2=0;
[c,d]=solve('c=-(q0-qp1-h*(d-ti))/k','-k*(-qi*l/k+c)=h*(-(qi/(2*k))*l^2+c*l+d-ti)-qp2');
c=subs(c,{'q0','qp1','h','ti','k','qp2','qi'},{q0,qp1,h,ti,k,qp2,qi})
d=subs(d,{'q0','qp1','h','ti','k','qp2','qi'},{q0,qp1,h,ti,k,qp2,qi})
这是正确代码,
c =
- (12*c)/5 - 5*c*l
d =
(6*c)/25 + d + c*l
结果如上.
给分吧,好的话,加送不介意,
c =
1/2*(-4*c*k^2+2*qp1*k+2*k*qi*l+h*qi*l^2-2*h*c*l*k+2*qp2*k)/k
d =
-1/2/h/k*(2*k*qi*l-2*c*k^2+h*qi*l^2-2*h*c*l*k-2*h*d*k+2*qp2*k)
没有问题!