how to solve one variable in non linear equation?

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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)

Accepted Answer

Torsten
Torsten on 12 Jan 2023
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

More Answers (1)

VBBV
VBBV on 12 Jan 2023
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 Comments
VBBV
VBBV on 12 Jan 2023
Read about fsolve for more info. Try with different initial values. I have shown an e.g.

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