# desolve_system where boundary values are functions of the dependent variable

 0 I am using desolve_system() to solve a boundary value problem. The equations are very simple but I'm having a bit of difficulty trying to include the boundary conditions. The system of equations and boundary conditions are, This is the code that I'm using in sage. How can I include the boundary values in this problem? reset() var('I_1 I_2 I0 R0 R1 x alpha D') I_1 = function('I_1', x) I_2 = function('I_2', x) eq1 = diff(I_1,x) == -alpha * I_1 eq2 = -diff(I_2,x) == -alpha * I_2 sol = desolve_system( [eq1, eq2], [I_1, I_2], ivar=x) view(sol) # How can I include these? #BV1: I_1(x=0)==A*I_2(x=0) + C #BV2: I_2(x=D)==B*I_1(x=D)  asked Jun 06 '12 Daniel Farrell 3 ● 3

 2 reset() var('I_1 I_2 I0 A B C D c c1 c2 x alpha D') I_1 = function('I_1', x) I_2 = function('I_2', x) eq1 = diff(I_1,x) == -alpha * I_1 eq2 = -diff(I_2,x) == -alpha * I_2 sol1 = desolve( eq1,I_1,ivar=x) sol2 = desolve( eq2,I_2,ivar=x) sol1=sol1.subs(c=c1) sol2=sol2.subs(c=c2) e1=sol1.subs(x=0)==A*sol2.subs(x=0)+C e2=sol2.subs(x=D)==B*sol1.subs(x=D) sol=solve([e1,e2],[c1,c2]) print "I_1 =",sol1.subs(sol[0][0]) print "I_2 =",sol2.subs(sol[0][1]) #I_1 = -C*e^(2*D*alpha - alpha*x)/(A*B - e^(2*D*alpha)) #I_2 = -B*C*e^(alpha*x)/(A*B - e^(2*D*alpha))  posted Jun 06 '12 achrzesz 1711 ● 4 ● 17 ● 38 Thank you very much, that's a very clever solution.Daniel Farrell (Jun 06 '12)

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