another function for solving differential equation other than dsolve in MATLAB

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I am using dsolve function in MATLAB to solve a differential equation. It is so slow and can not end up with a result. Is there a way to speed it up, or is there another way for solving a differential equation? I don't want to solve it numerically.
The equations are as follow:
all the derivatives in above equations are replaced with equivalent expressions, so all are only function of y and T only.
I need to solve this equation:
Cg, Cs and qi are constants. The initial conditions for positive root is (y=0.01, T=0) and for negative root is (y=0.99, T=0).
Here is my code:
R=1.9872; %cal/mol/K
eps=0.763;
rhos=0.484; %gr/cm3
T0=300; %theta K
y0=0.01;
ys=0.99;
p=1; %atm
Upg=0.68e-3; %mol/cm2/s
L=100; %cm column length
Cg=1.04*239.006/1000*28.0134; % kj/kg/k to cal/mol/k N2
Cs=0.22; %cal/gr/k
% m(mol/gr) b(1/atm) q(cal/mol)
cA=[3.65e-3 2.8e-4 4900]; % co2 BPL AC
cB=[3.65e-3 13.5e-4 2500]; % N2 BPL AC
i=1;
IPA=cA(i,:); IPB=cB(i,:);
IPA1=IPA(1);IPA2=IPA(2);IPA3=IPA(3);
IPB1=IPB(1);IPB2=IPB(2);IPB3=IPB(3);
syms T(y)
DN1T= @(y,T) (IPA1*IPA2*p*y*exp(IPA3/(R*(T + T0)))*((IPA2*IPA3*p*y*exp(IPA3/(R*(T + T0))))/(R*(T + T0)^2) - (IPB2*IPB3*p*exp(IPB3/(R*(T + T0)))...
*(y - 1))/(R*(T + T0)^2)))/(IPA2*p*y*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1)^2 -...
(IPA1*IPA2*IPA3*p*y*exp(IPA3/(R*(T + T0))))/(R*(T + T0)^2*(IPA2*p*y*exp(IPA3/(R*(T + T0))) - ...
IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1));
DN2T =@(y,T) (IPB1*IPB2*IPB3*p*exp(IPB3/(R*(T + T0)))*(y - 1))/(R*(T + T0)^2*(IPA2*p*y*exp(IPA3/(R*(T + T0))) - ...
IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1)) - (IPB1*IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1)*((IPA2*IPA3*p*y*...
exp(IPA3/(R*(T + T0))))/(R*(T + T0)^2) - (IPB2*IPB3*p*exp(IPB3/(R*(T + T0)))*(y - 1))/(R*(T + T0)^2)))...
/(IPA2*p*y*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1)^2;
DN1Y =@(y,T) (IPA1*IPA2*p*exp(IPA3/(R*(T + T0))))/(IPA2*p*y*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1)...
- (IPA1*IPA2*p*y*exp(IPA3/(R*(T + T0)))*(IPA2*p*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))))/...
(IPA2*p*y*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1)^2;
DN2Y =@(y,T) (IPB1*IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1)*(IPA2*p*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))))/...
(IPA2*p*y*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1)^2 - (IPB1*IPB2*p*exp(IPB3/(R*(T + T0))))...
/(IPA2*p*y*exp(IPA3/(R*(T + T0))) - IPB2*p*exp(IPB3/(R*(T + T0)))*(y - 1) + 1);
A=@(y,T) Cg*(DN1T(y,T)-y*(DN1T(y,T)+DN2T(y,T)));
B=@(y,T) (Cg*(T)*(DN1T(y,T)+DN2T(y,T))-Cs)+Cg*(DN1Y(y,T)-y*(DN1Y(y,T)+DN2Y(y,T)))+cA(i,3)*DN1T(y,T)+cB(i,3)*DN2T(y,T);
C=@(y,T) Cg*(T)*(DN1Y(y,T)+DN2Y(y,T))+cA(i,3)*DN1Y(y,T)+cB(i,3)*DN2Y(y,T);
S4=@(y,T) (-B(y,T) +(B(y,T) ^2-4*A(y,T) *C(y,T) )^0.5)/2/A(y,T) ;
S2=@(y,T) (-B(y,T) -(B(y,T) ^2-4*A(y,T) *C(y,T) )^0.5)/2/A(y,T) ;
S4sol=dsolve([diff(T,y)==S4(y,T) , T(y0)==0])
S2sol=dsolve([diff(T,y)==S2(y,T) , T(ys)==0])
  5 个评论
Bjorn Gustavsson
Bjorn Gustavsson 2020-10-5
No, I mean write the equations like you would see them in a book or article:
That makes it far easier for others to read than your long lines of code.

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采纳的回答

Kiran Felix Robert
Kiran Felix Robert 2020-10-8
Hi Leila,
One way to speed-up the execution is to add a limit on the maximum degree of radicals. The following shows you an example,
S = dsolve(..,'MaxDegree',2);
This will force dsolve to assume implicit formulas for polynomials of degree greater than the specified value.
Refer to the dsolve documentation for more information.
Kiran Felix Robert

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