Required to solve a MATRIX of DE`s, using Taylor series. Form X` =...

Required to solve a MATRIX of DE`s, using Taylor series. Form X` = AX, where X` and X are vectors and A is coefficient matrix ( which may be functions - but in this case not )

A = [1 1 4; 0 2 0; 1 1 1] and X(0) = [ -2 0 1 ] T. ( A in rows and x(0) transpose )

Use the power series method to find a solution for this equation.

NOTE : this is NOT A simple series solution methodology of a single ODE. If one uses the eigenvalue/eigenvector method one finds eigenvalues of -1, 2, & 3 with corresponding eigenvectors of [ -2 0 1 ] T ; [ 5 -3 2] T & [ 2 0 1] T respectively. So solution function is EV1 * exp ( Lambda1 ) + EV2*exp (Lambda2 ) + EV3*exp( Lambda3 ) in normal fashion (lambda`s and corresponding EV`s provided ). So we know the answer!

But the trick is to develop the matrix A in multiples of A times X(0) in order to discern a pattern . Then to describe that pattern in terms of a Taylor series. The structure we are looking for in this case is SIGMA (for capital summation function) k = 0 to infinity of 1/k!* (t **k ), - which corresponds to the EXP function. ( and we know this is the Taylor form of the exponential function we got from the eigenvector solution given)

Key equations are X(t) = X(0) + SIGMA ( k = 0 to infinity ) 1/k! * (A**k)* X(0)*(t**k), and also X(n) = 1/n! *(A**n)*X(0) - because A is a matrix of constants. So A*X(0), A**2*X(0), A**3*X(0) etc need to be calculated manually in each case, in order to look for a pattern in the resultant vector that repeats in some fashion. Then describe this repeating pattern in terms of an infinite ( Taylor ) series. Then fiddle around with the algebra of the resultant series in order to see if one can match up exactly to the Taylor series format of the eigenvector type solution. Then, one has proved that there is an answer using infinite Taylor series that matches the eigenvalue derived solution.

REF: Systems of ODE`s, Goldberg & Schwartz , Harper & Row, 1972, PP 194 - 209.