Solving First order ODEs simultaneously
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Hello, needed help figuring out why I cannot obtain a solution. I'm sure this is a solvable solution however I keep getting a warning saying no solution is found. Is there any mistake I'm making in the code?
Everything is a constant except E, Sr(t) & Er(t).
% Rigorous Solution Case #1
syms Sr(t) Er(t) E;
E = Ea - Er(t);
ode2a = diff(Sr(t),t) == -(k1*(Ea - Er(t))*Sr(t)) + krev1*Er(t);
ode3a = diff(Er,t) == (k1*(Ea - Er(t))*Sr(t)) - (krev1+k2)*Er(t);
odes = [ode2a; ode3a];
cond1 = Sr(0) == Sa;
cond2 = Er(0) == 0;
conds = [cond1; cond2];
[SrSol(t),ErSol(t)] = dsolve(odes,conds)
4 Comments
Ea = 123; %just to have SOME value
k1 = 42; %just to have SOME value
k2 = 13; %just to have SOME value
krev1 = 48; %just to have SOME value
Sa = 5; %just to have SOME value
% Rigorous Solution Case #1
syms Sr(t) Er(t) E;
E = Ea - Er(t);
ode2a = diff(Sr(t),t) == -(k1*(Ea - Er(t))*Sr(t)) + krev1*Er(t);
ode3a = diff(Er,t) == (k1*(Ea - Er(t))*Sr(t)) - (krev1+k2)*Er(t);
odes = [ode2a; ode3a];
cond1 = Sr(0) == Sa;
cond2 = Er(0) == 0;
conds = [cond1; cond2];
sol = dsolve(odes,conds)
sol2a = dsolve(ode2a)
so3a = dsolve(ode3a)
I don't know that I would call those "solvable"
Valerie
on 28 Sep 2023
Ea = 123; %just to have SOME value
k1 = 42; %just to have SOME value
k2 = 13; %just to have SOME value
krev1 = 48; %just to have SOME value
Sa = 5; %just to have SOME value
% Rigorous Solution Case #1
syms Sr(t) Er(t) E;
E = Ea - Er(t);
ode2a = diff(Sr(t),t) == -(k1*(Ea - Er(t))*Sr(t)) + krev1*Er(t);
ode3a = diff(Er,t) == (k1*(Ea - Er(t))*Sr(t)) - (krev1+k2)*Er(t);
eqns = [ode2a; ode3a];
cond1 = Sr(0) == Sa;
cond2 = Er(0) == 0;
conds = [cond1; cond2];
[eqs,vars] = reduceDifferentialOrder(eqns, [Sr(t), Er(t)])
[M,F] = massMatrixForm(eqs,vars)
f = M\F
odefun = odeFunction(f,vars)
InitConditions = double(rhs(conds)) %watch out for order though!
[T, Y] = ode45(odefun, [0 0.01], InitConditions);
subplot(2,1,1); plot(T, Y(:,1)); title(string(vars(1)))
subplot(2,1,2); plot(T, Y(:,2)); title(string(vars(2)))
%that almost looks like the initial conditions are reversed.
%what happens if we try reversing the conditions?
[Tr, Yr] = ode45(odefun, [0 0.01], flipud(InitConditions));
figure
subplot(2,1,1); plot(Tr, Yr(:,1)); title(string(vars(1)))
subplot(2,1,2); plot(Tr, Yr(:,2)); title(string(vars(2)))
Valerie
on 28 Sep 2023
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