clear

%%
% We'll examine the guessing/IVP approach for the case lambda=0.6 
lambda = 0.6;
phi = @(r, w, dwdr) lambda./w.^2 - dwdr./r;

%%
a = eps;  % We have to avoid r=0, or we would divide by zero.
b = 1;

%%
% We convert the ODE to a first-order system in v1, v2. 
f = @(r,v) [ v(2); phi(r, v(1), v(2)) ];

%%
% Now we try multiple guesses for the unknown w(0) and plot the
% solutions. 
for w0 = 0.5:0.1:0.9
    [r, v] = ode45(f, [a, b], [w0; 0]);
    plot(r, v(:,1))
    hold on
end
xlabel('x')
ylabel('w(x)')
title('Solutions for choices of w(0)')
grid on
legend('w(0) = 0.5', 'w(0) = 0.6', 'w(0) = 0.7', 'w(0) = 0.8', 'w(0) = 0.9', 'location', 'best')

%%
% On the graph, it's the curve starting at $w(0)=0.8$ that comes closest
% to the required condition $w(1)=1$.