How do I get the second solution?

function Shrink;
format short
global lambda Pr 
Pr = 1;  %fixed
a = 0;
b = 10;  % b = infinity
lambda =-1.1; %mixed convection parameter (lambda)
solinit = bvpinit(linspace(a,b,100),@Shrink_init);
options = bvpset('stats','on','RelTol',1e-10);
sol = bvp4c(@Shrink_ode,@Shrink_bc,solinit,options);
figure(1)
plot(sol.x,sol.y(2,:),'k--','linewidth',1)
xlabel('\eta','fontsize',16)
ylabel('f\prime(\eta)','fontsize',16)
hold on
figure(2)
plot(sol.x,sol.y(4,:),'k--','linewidth',1)
xlabel('\eta','fontsize',16)
ylabel('\theta(\eta)','fontsize',16)
hold on
sol.y(3,1)  % The value of f''(0)
-sol.y(5,1) % The value of -theta'(0)
descris = [sol.x; sol.y];
save ('start_main.mat', '-struct', 'sol');
function dydx = Shrink_ode(x,y,Pr,lambda);
global Pr lambda
dydx = [y(2)
    y(3)
    y(2)^2-y(1)*y(3)-1-lambda*y(4)
    y(5)
    -Pr*y(1)*y(5)+Pr*y(2)*y(4)];


function res = Shrink_bc(ya,yb);
res = [ya(1)
    ya(2)
    yb(2)-1
    ya(4)-1
    yb(4)];

function v = Shrink_init(x);
v =[];???????????????

[Jan]

The question is not clear. Please take the time to explain any details.

Answer 1

[Torsten]

Pr = 1;  %fixed
lambda =-1.1; %mixed convection parameter (lambda)
a = 0;
b = 10;  % b = infinity

solinit = bvpinit(linspace(a,b,100),@Shrink_init);
options = bvpset('stats','on','RelTol',1e-10);
sol = bvp4c(@(x,y)Shrink_ode(x,y,Pr,lambda),@Shrink_bc,solinit,options);

Warning: Unable to meet the tolerance without using more than 2000 mesh points.
The last mesh of 887 points and the solution are available in the output argument.
The maximum residual is 2.0419e-06, while requested accuracy is 1e-10.

The solution was obtained on a mesh of 2631 points.
The maximum residual is  2.042e-06. 
There were 55777 calls to the ODE function. 
There were 226 calls to the BC function. 

figure(1)
plot(sol.x,sol.y(2,:),'k--','linewidth',1)
xlabel('\eta','fontsize',16)
ylabel('f\prime(\eta)','fontsize',16)

figure(2)
plot(sol.x,sol.y(4,:),'k--','linewidth',1)
xlabel('\eta','fontsize',16)
ylabel('\theta(\eta)','fontsize',16)

%hold on
sol.y(3,1)  % The value of f''(0)

ans = -0.2153

-sol.y(5,1) % The value of -theta'(0)

ans = -0.1098

%descris = [sol.x; sol.y];
%save ('start_main.mat', '-struct', 'sol');
function dydx = Shrink_ode(x,y,Pr,lambda);
dydx = [y(2)
    y(3)
    y(2)^2-y(1)*y(3)-1-lambda*y(4)
    y(5)
    -Pr*y(1)*y(5)+Pr*y(2)*y(4)];
end


function res = Shrink_bc(ya,yb);
res = [ya(1)
    ya(2)
    yb(2)-1
    ya(4)-1
    yb(4)];
end

function v = Shrink_init(x);
v =[0 1 0 1 1];
end

[MOSLI KARIM]

Hi Mr. Torsten, I have a question, why did you choose the initial conditions

function v = Shrink_init(x);
v =[0 1 0 1 1];
end

[Torsten]

I took one of the boundary conditions as initial value for the complete region of integration.

Somewhat arbitrary - but often it works.