How to code an equation with that integrates a vector.

Answer 1

[Walter Roberson]

There are two ways.

You can do it symbolically:

syms A B C
syms u__vec_(t) [1 5]
x_0__vec = [A*cos(B)*cos(C)
            5 + A*sin(B)
            A*cos(B)*sin(C)];
x__vec = x_0__vec + int(u__vec_, t, 0, t)

x__vec = 

Or you can do it numerically

An = 1.2; Bn = pi/8; Cn = 2*pi/5;
x0_vec = [An*cos(Bn)*cos(Cn)
            5 + An*sin(Bn)
            An*cos(Bn)*sin(Cn)];
tn = 5;
u_vec = @(t) [t.^2/5, exp(-t), sqrt(t)];
x_vec = x0_vec + integral(u_vec, 0, tn, 'arrayvalued', true)

x_vec = 3×3

8.6759 1.3359 7.7962 13.7926 6.4525 12.9128 9.3877 2.0477 8.5080

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Note that in the numeric case, you must define a specific numeric upper-bound for the integration.

Answer 2

[Star Strider]

Siince ‘u’ is also a vector —

u = @(t) [sinh(t); cosh(t); tanh(t)]

u = function_handle with value:
    @(t)[sinh(t);cosh(t);tanh(t)]

int_u = integral(u, 0, 1, ArrayValued=true)

int_u = 3×1

0.5431 1.1752 0.4338

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A = rand

A = 0.9710

B = rand

B = 0.3265

C = rand

C = 0.6471

x0 = [A.*cos(B).*cos(C); 5+A*sin(B); A.*cos(A).*sin(C)]

x0 = 3×1

0.7338 5.3114 0.3304

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●

xv = x0 + integral(u, 0, 1, ArrayValued=true)

xv = 3×1

1.2768 6.4867 0.7642

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Make appropriate changes to ‘u’ and the constants to get the result you want.

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Answer 3

[Torsten]

If there is no closed-form expression for u (e.g. if u depends on x), you can solve the vector-differential equation

dx/dt = u
x(0) = x0