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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 | --Copyright The Numerical Algorithms Group Limited 1994.
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-- There are 2 different ways to input a homogeneous 1st order system of
-- linear ordinary differential equations of the form dy/dt = M y
-- where y is a vector of unknown functions of t.
-- the first is simply solve(M, t) which will be understood to be
-- a differential system:
M := matrix [[ 1+4*t, -5*t, 7*t, -8*t, 8*t, -6*t],_
[ -10*t, 1+9*t, -14*t, 16*t, -16*t, 12*t],_
[ -5*t, 5*t, 1-8*t, 8*t, -8*t, 6*t],_
[ 10*t, -10*t, 14*t,1-17*t, 16*t, -12*t],_
[ 5*t, -5*t, 7*t, -8*t, 1+7*t, -6*t],_
[ -5*t, 5*t, -7*t, 8*t, -8*t, 1+5*t]]
-- the original system in Barkatou's AAECC paper is t^2 dy/dt = M*y
sol := solve(inv(t**2) * M, t)
-- verify the solutions
[t**2 * map(h +-> D(h, t), v) - M * v for v in sol]
-- the second way is to type each equation using a separate operator for
-- each unknown:
x := operator x
y := operator y
sys := [D(x t, t) = x t + sqrt 3 * y t, D(y t, t) = sqrt 3 * x t - y t]
solve(sys, [x, y], t).basis
-- Similarly there are 2 different ways to input the inhomogeneous system
-- dy/dt = M y + v where v is a given vector of functions.
-- the first is solve(M, v, t):
v := vector [1, (-29*t + 19)/5, -1, t + 1, - 2*t + 3, -1]
-- get a particular solution to t^2 dy/dt = M y + v
solp := solve(inv(t**2) * M, inv(t**2) * v, t).particular
-- verify the particular solution
t**2 * map(h +-> D(h, t), solp) - M * solp - v
-- the second way is by listing the equations:
z := operator z
sys := [D(x t, t) = y t + z t + t, D(y t, t) = x t + z t, D(z t, t) = x t + y t]
solve(sys, [x, y, z], t).particular
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