# How to solve first order differential condition with two initial conditions?

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I G on 27 Feb 2020
Commented: darova on 27 Feb 2020
I had system of equations defined in m file fun_zap_ri.m:
function [f, R] = fun_zap_ri(z, p, beta, ri)
R = ri - z .* (ri - 1);
f = zeros(4, size(p,2));
f(1,:) = - 32 .* beta ./ (R .^ 4 .* p(1,:));
f(2,:) = ( - 8 .* f(1,:) ./ R - f(1,:) .* p(2,:) ) ./ p(1,:);
f(3,:) = ( - p(2,:) .* f(2,:) - 8 .* f(2,:) ./ R - 8 .* f(1,:) ./ (R .* R .* p(1,:) ) - f(1,:) .* p(3,:) ) ./ p(1,:);
f(4,:) = ( - f(2,:) .* p(3,:) - f(3,:) .* p(2,:) + 8 .* ( - f(3,:) ./ R - ( f(2,:) ./ p(1,:) - p(2,:) .* f(1,:) ./ ( p(1,:) .* p(1,:) ) ) ./ (R .* R)) - f(1,:) .* p(4,:) ) ./ p(1,:);
where I solved it with call:
ri = 1;
beta = 1;
z = linspace(1,0,1001);
options = odeset('RelTol',1.e-6);
[~, pv] = ode45(@(z, p)fun_zap_ri(z, p, beta, ri), z, [1; 0; 0; 0], options);
The system given above I got from form of equation (I will give example just for the first equation)
p0 * p0' = - 32 * beta * m0 / R^4
where m0 = 1, than it can be concluded p0' = - 32 * beta * m0 / ( R^4 * p0)
I also have condition p0(z = 1) = 1 (exactly this is part of code given in [1; 0; 0; 0]). With all other given parameters, as z, ri, R, beta I can calculate p0 (or as given in my code pv) with ode45 call.
Now I have the same system p0 * p0' = - 32 * beta * m0 / R^4, but m0 is not given, but I have values for p0(z=0) = 8 and p0(z=1) = 1.
How I can solve this differential equation? Should I also give m0? Or is there some function in matlab which will solve equation with two "initial" conditions?
darova on 27 Feb 2020
I think bvp4c should work for this case

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