Symbolically solve non-linear differential equation

I need to solve this equation symbolically because I need to use in integration

c1=0.047;
c2=0.053;
m1=0.0025;
m2=0.008;
l1=1/500;
l2=1/20;
b10=0.1;
b20=0.2;
syms t;
syms b2(t)

b1t=-(c1-m1)*b10/(((b10-1)*c1+m1)*exp(-(c1-m1)*t)-c1*b10);
eq2=diff(b2,t)==b2(t)*(c2*(1-b1t-b2(t))-m2-c1*b1t);

b2t=dsolve(eq2,b2(0)==b20);

num=eval(vpaintegral((l1*b1t+l2*b2t)*t*exp(-(l1*b1t+l2*b2t)*t),[0 10000]));

c, m and l are fixed paramters, b10 and b20 change for different setups of the simulation. When b10 is under 0.05 b2t can be computed in 10-15 seconds, but starting around b10=0.1 I cannot get it to work. I have let it run for over 40 minutes with no answer.

I know it has a solution because Maple can solve it in seconds, but all my code is in matlab and I cannot just change it now. I can't just copy it because the expresion is infernally long. Also I can evaluate it with ode45 with no problems, but then I don't have an expresion to integrate...

Does anyone know if there are any options to dsolve that I can use or maybe another speciallised package/function for these situations?

Thank you and have a good day.

Answer 1

[Torsten]

You don't need to solve the equation symbolically.

Just solve simultaneously the two differential equations

%db2/dt=b2*(c2*(1-b1t-b2)-m2-c1*b1t) , b2(0) = b20
%dI/dt = (l1*b1t+l2*b2t)*t*exp(-(l1*b1t+l2*b2t)*t), I(0) = 0

by a numerical method (e.g. using ODE15S).

The value of I at t = 10000 gives you the integral value you are after.

c1=0.047;
c2=0.053;
m1=0.0025;
m2=0.008;
l1=1/500;
l2=1/20;
b10=0.2;
b20=0.2;
b1t=@(t)-(c1-m1)*b10/(((b10-1)*c1+m1)*exp(-(c1-m1)*t)-c1*b10);
fun = @(t,y)[y(1)*(c2*(1-b1t(t)-y(1))-m2-c1*b1t(t));(l1*b1t(t)+l2*y(1))*t*exp(-(l1*b1t(t)+l2*y(1))*t)];
tspan = [0 10000];
y0 = [b20;0];
[T,Y] = ode15s(fun,tspan,y0,odeset('RelTol',1e-12,'AbsTol',1e-12));
plot(T,Y(:,1))

Y(end,2)

ans = 542.0010

[Pedro]

Thanks!

I had no idea that could be done. I managed to make it run in the end by integrating with b1t as symbolic as well and using subs with the final expression. I will see which way runs faster, since I need to loop this calculation.

Pedro

[Torsten]

I personally see no sense in making b1t symbolic apart from making your code slower, but maybe you have your reasons.

[Pedro]

I was just traying things and it happened to help solving b2t, where my code was getting completely stuck.