PLEASE help : system of ODE

I have system of five differential equations with initial conditions. They are as follows. Kindly help me solve them.

Dx=((((0.4*a)/(a*(0.029*x)))*(0.668/(b+0.668)))-((1/v)*0.01))*x

Da=((0.01/v)*50)-(((((2/5)+(13*0.13)+(3*0.051)+(3*2.62))*(0.01/v))+(3*0.188))*x)-((0.01/v)*a)

Db=(2*(((0.051+(2*(0.13-(1/20)))+2.62)*(0.1/v))+0.188))-((0.01/v)*b)

Dc=((6*(0.13-(1/20)))*(0.01/v))-((0.01/v)*c)

Dv=0.01

the initial conditions are as follows:

x(0)=4.41

a(0)=0

b(0)=22.68

c(0)=1.28

v(0)=1

Answer 1

[Friedrich]

Hi,

Based on this article (chapter 2.3)

http://www.math.tamu.edu/REU/comp/matode.pdf

I think you have to do it like this:

function my_func()

start_cond = [4.41, 0, 22.68, 1.28, 1];
tspan = [0,20];
[t,x] = ode45(@func,tspan,start_cond);
plot(t,x);

function out = func(t,x)

%x(1) is variable x
%x(2) is variable a
%x(3) is variable b
%x(4) is variable c
%x(5) is variable v

out = zeros(size(x));

out(1) = ((((0.4*x(2))/(x(2)*(0.029*x(1))))*(0.668/(x(3)+0.668)))-((1/x(5))*0.01))*x(1);
out(2) = ((0.01/x(5))*50)-(((((2/5)+(13*0.13)+(3*0.051)+(3*2.62))*(0.01/x(5)))+(3*0.188))*x(1))-((0.01/x(5))*x(2));
out(3) = (2*(((0.051+(2*(0.13-(1/20)))+2.62)*(0.1/x(5)))+0.188))-((0.01/x(5))*x(3));
out(4) = ((6*(0.13-(1/20)))*(0.01/x(5)))-((0.01/x(5))*x(4));
out(5) = 0.01;

end

Please double check the code for typos in regard of the x(1,...5) convertion!

[Arun Maari Rajha K V]

Thanks for your reply sir. However ode45 solver helps in finding the function values at a particular time span. Is there any way to get the general solution i.e. I mean the each function's expression as a function of other dependent variables.

Thanks.

[Friedrich]

When you like to get a closed solution you need the Symbolic Math Toolbox which contains the function dsolve:

http://www.mathworks.com/help/releases/R2011a/toolbox/symbolic/dsolve.html

[Arun Maari Rajha K V]

could you kindly show me how to do it sir?

[Friedrich]

Acording to the doc I would say like this:

out = dsolve('Dx=((((0.4*a)/(a*(0.029*x)))*(0.668/(b+0.668)))-((1/v)*0.01))*x',...

'Da=((0.01/v)*50)-(((((2/5)+(13*0.13)+(3*0.051)+(3*2.62))*(0.01/v))+(3*0.188))*x)-((0.01/v)*a)',...

'Db=(2*(((0.051+(2*(0.13-(1/20)))+2.62)*(0.1/v))+0.188))-((0.01/v)*b)',...

'Dc=((6*(0.13-(1/20)))*(0.01/v))-((0.01/v)*c)',...

'Dv=0.01','x(0)=4.41','a(0)=0','b(0)=22.68','c(0)=1.28','v(0)=1')

But I get an error which I don't understand. Than I tried it without the initial conditions:

out = dsolve('Dx=((((0.4*a)/(a*(0.029*x)))*(0.668/(b+0.668)))-((1/v)*0.01))*x',...

'Da=((0.01/v)*50)-(((((2/5)+(13*0.13)+(3*0.051)+(3*2.62))*(0.01/v))+(3*0.188))*x)-((0.01/v)*a)',...

'Db=(2*(((0.051+(2*(0.13-(1/20)))+2.62)*(0.1/v))+0.188))-((0.01/v)*b)',...

'Dc=((6*(0.13-(1/20)))*(0.01/v))-((0.01/v)*c)',...

'Dv=0.01')

which worked fine. The result for b,c,v looks good. x and a looking very complicated. You have to calculate the unknown C** variables afterwards. Since you have the initial conditions this should be possible but totally ugly.

Why not using the numerically solution? Its faster and easier to handle.