Calculus Variational; code for finding the value of two constants for 2 values of the variable

Hello,

I want to write a code for Calculus Variational problem. I have to find the extremals this equation J = $(0->1) [diff(x,t)^2+10*t*x]dt

After the Euler equation, I found x(t) == (5*t^3)/6 + c1*t + c2; but here I didn't suceed in writing the code for calculating c1 and c2 for t=0 and t=1.

Is there a chance for anyone to help me?

Thank you,

% optimal control course
%Example: optimal control problem, min J = $(0->1) [diff(x,t)^2+10*t*x]dt
%boundary conditions: x(0) = 1, x(1)=2;
%dsolve – solve differential equations
%solve – solve algebraic equations
%solve differential equations (Euler – Lagrange equations)
syms x(t)
f = diff(x,t)^2+10*t*x;
G = functionalDerivative(f,x)
%solve differential equations (Euler – Lagrange equations)
syms x(t)
eqns =[ x(t) == (5*t^3)/6 + c1*t + c2];
%solve algebraic equations by applying boundary conditions
syms t0 tf
t0=0
tf=1
SS = solve(eqns);
c1 = SS.c1;
c2 = SS.c2;

Answer 1

[Paul]

Boundary conditions can be specified in the call to dsolve

syms x(t) 
f = diff(x,t)^2+10*t*x

f(t) = 

eqn = functionalDerivative(f,x) == 0

eqn(t) = 

sol = dsolve(eqn,x(0)==1,x(1)==2)

sol = 

[Viorica Alina]

Thank you very much; this is perfect!!!

[Dyuman Joshi]

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Answer 2

[Torsten]

I didn't check whether the Euler equation gives x(t) = 5*t^3/6 + c1*t + c2, but if this is the case, you can get c1 and c2 via

syms t c1 c2
x(t) = 5*t^3/6 + c1*t + c2;
eqn1 = x(0)==1;
eqn2 = x(1)==2;
sol = solve([eqn1,eqn2],[c1,c2]);
xsol = subs(x,[c1,c2],[sol.c1,sol.c2])

xsol(t) = 

[Viorica Alina]

Thank you; very helpful!