(1+2M*eta)f''' + 2M*f"+ f*f" -f'^2 – k1*f' + lambda*theta=0 ———- (1)
(1+2M*eta)theta" + 2M*theta' + Pr(f*theta'-f'*theta)=0 ——-(2)
'f' and 'theta' are functions of 'eta', eta is an independent variable
3 initial conditions are given: eta=0, f(0)=0, f'(0)=1,theta(0)=1
Say I reduce these equations (1) and (2) to five ode (shooting method)
f'=z ; f(0)=0 —–(3)
z'=p ; z(0)=1 ——(4)
p'= (-2M*f"-f*f"+f'^2+k1*f'-lamda*theta)/(1+2*M*eta) ;p(0)= (guess value) —–(5)
theta'= q ; theta(0)=1 —–(6)
q' = (-2M*theta'-Pr(f*theta'-f'*theta))/(1+2*M*eta) ; q(0)= (guess value) ——(7)
The boundary conditions that needs to be satisfied are: f'(eta=10)= 0 and theta(eta=10)=0 as eta=10
Given:
M= 1
k1= 0.1
lamda= 0.1
Pr= 0.7
taking step length: h= 0.01
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