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You don't even need indices. If you have defined your matrix (A) and vector (V), just store your result in vector B=A*V. Thus, if A is (2x2) and V is (2x1) the resulting vector B is 2x1.

I've discovered that Odesolve can solve this kind of DE, however it won't converge to a solution, unless the boundary becomes dy/dx=(2x)^-0.5 at x=2. If x>2 no solution is possible.

Thank Occupant. This is exactly what I wanted. I realized what my my mistake was. Thanks a lot. If you like, take a look at my other question about the nonlinear DE.

I got a simple system of two diff. eqs; The functions in play are: z(t)=sin(t) and w(t)=cos(t). Tried to solve the system Given z'(t) = w(t) w'(t) = -z'(t) with IC: z'(0)=1 w'(0)=0 S=Odesolve(t,8?,100) Solution fails... Even tried Rkadapt. Any ideas; (See attached MC file)..

Hi! I'm currently struggling to solve the following nonlinear 2nd order DE with MC: y'' - (1/y)(y')^2 - y(dy/dx + 1) = 0 with ic: y(0)=0.235 and bc: dy/dx=(2x)^-0.5 at x=50. I've tried odesolve and rkadapt with no success. In fact, everywhere I looked I got solutions for ODE only. Can't...

Right! However, it sounds complicated seeking out roots when the equation is in matrix form. I could try this in the future. Thanks again.

Thanks, IRstuff. Your solution is right. However, I am not going to use it after all. The plot should produce an ellipse (joint confidence region of the two parameters), but due to the fact that MC returns one only root from the quadratic equation, the corresponding plot shows one only branch of...

Thanks for the answers, but it doesn't work. The problem is that I cannot make one of the elements of vector v a range variable, since MC won't allow this. To give you a better idea I rewrite my block in detail: Vo=1 v1=1 (initial guesses) Given (v-b)'(X'X)(v-b) = F Find(v) Now, b is a...

Hi I am trying to iterate a solution from a solve block, but nothing seems to work. My block contains a simple matrix equation yielding a solution for the vector v (of two elements only), i.e., Solve (v-b)'(X'X)(v-b)=G Find(v) Now, v consists of v0 and v1 elements. I want to use...