how to solve one variable in non linear equation?

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M.Rameswari Sudha
M.Rameswari Sudha 2023-1-12
评论： M.Rameswari Sudha 2023-1-12
In my coding I couldn't get the value for t1. I get the answer in cubic equation of another variable z. But I didn't use z anywhere in my codind. Anyone help and tell how to get the answer for t1.
clc
clear all
T=12; u1=4;u2=8;a=30;b=5;a2=100;A=500;c2=10;c3=12;c4=8;D0=115; b2=0.2;
a=0.01;d=0.2;m=0.5;k1=0.5;k0=1;k2=2;
syms t1
c1=5;
h=(1./T).*(c1.*((b-a)-b.*t1-(a+b.*(1+m)./(1+m-t1)))+c2.*(b.*t1-(a+b.*(1+m)).*(t1./(1+m-t1)))+c3.*((a./k1)-((b./k1.^2).*(1-k1.*T)).*(((-u1.*k1)./(1+k1.*(t1-T))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+(b.*k0.*(t1-u1)./k1)-(a-(b.*k0.*(1-k1.*T)./k1)).*((u2-u1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(u1-u2)./(1+k1.*(t1-T)))-(a-(b.*k0./k1).*(1-k1.*T)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0./k1).*(u2-T)-(1-k1.*T).*((T+u2)./(1+k1.*(t1-T)))+c4.*(-a2+b2.*t1+k0.*(a-(b./k1).*(1-k1.*T).*(1./(1+k1.*(t1-T))+(b./k1))))))));
t1=solve(h)
disp(t1)

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采纳的回答

Torsten
Torsten 2023-1-12
Ran in:
T=12; u1=400;u2=8;a=30;b=5;a2=100;A=500;c2=10;c3=12;c4=8;D0=115; b2=0.2;
a=-1.1;d=11.2;m=-0.5;k1=1.5;k0=1.1;k2=2;
syms t1
c1=5;
h= (1./T).*(c1.*((b-a)-b.*t1-(a+b.*(1+m)./(1+m-t1)))+c2.*(b.*t1-(a+b.*(1+m)).*(t1./(1+m-t1)))+c3.*((a./k1)-((b./k1.^2).*(1-k1.*T)).*(((-u1.*k1)./(1+k1.*(t1-T))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+(b.*k0.*(t1-u1)./k1)-(a-(b.*k0.*(1-k1.*T)./k1)).*((u2-u1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(u1-u2)./(1+k1.*(t1-T)))-(a-(b.*k0./k1).*(1-k1.*T)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0./k1).*(u2-T)-(1-k1.*T).*((T+u2)./(1+k1.*(t1-T)))+c4.*(-a2+b2.*t1+k0.*(a-(b./k1).*(1-k1.*T).*(1./(1+k1.*(t1-T))+(b./k1))))))));
t1=vpasolve(h,t1,[0 1])
t1 = 
0.4999993236378365887378927054944
double(subs(h,t1))
ans = 2.2769e-24

更多回答（1 个）

VBBV
VBBV 2023-1-12
Ran in:
clc
clear all
T=12; u1=400;u2=8;a=30;b=5;a2=100;A=500;c2=10;c3=12;c4=8;D0=115; b2=0.2;
a=-1.1;d=11.2;m=-0.5;k1=1.5;k0=1.1;k2=2;
syms t1
c1=5;
h=@(t1) (1./T).*(c1.*((b-a)-b.*t1-(a+b.*(1+m)./(1+m-t1)))+c2.*(b.*t1-(a+b.*(1+m)).*(t1./(1+m-t1)))+c3.*((a./k1)-((b./k1.^2).*(1-k1.*T)).*(((-u1.*k1)./(1+k1.*(t1-T))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+(b.*k0.*(t1-u1)./k1)-(a-(b.*k0.*(1-k1.*T)./k1)).*((u2-u1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(u1-u2)./(1+k1.*(t1-T)))-(a-(b.*k0./k1).*(1-k1.*T)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0./k1).*(u2-T)-(1-k1.*T).*((T+u2)./(1+k1.*(t1-T)))+c4.*(-a2+b2.*t1+k0.*(a-(b./k1).*(1-k1.*T).*(1./(1+k1.*(t1-T))+(b./k1))))))));
t1=fsolve(h,[0 1])
Equation solved, solver stalled. fsolve stopped because the relative size of the current step is less than the value of the step size tolerance squared and the vector of function values is near zero as measured by the value of the function tolerance.
t1 = 1×2
10.4643 10.4643
disp(t1)
10.4643 10.4643
  4 个评论
显示 2更早的评论
M.Rameswari Sudha
M.Rameswari Sudha 2023-1-12
I want to solve t1 from h. No need to find h value. Help me
VBBV
VBBV 2023-1-12
Read about fsolve for more info. Try with different initial values. I have shown an e.g.

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