How to solve the system of such differential equations?

Question

Let $x_i=x_i(t)$ be a functions and let consider the system of differential equations

$\left \{ \begin{align} &{\frac { d}{{ d}t}}x_{{0}} =3\,x_{{1}} ,\\ &{\frac { d}{{ d}t}}x_{{1}} =0,\\ &{\frac { d}{{ d}t}}x_{{2}} = {\frac {2\,x_{{1}} x_{{2}} +x_{{3}} }{x_{{0}} }},\\ &{\frac { d}{{ d} t}}x_{{3}} ={\frac {3\,x_{{1}} x_{{3 }} -6\, x_{{2}} ^{2} }{x_{{0}} }}. \end{align} \right. $

My attempt. By brute forse I have found the first integral

$$ {\frac {4\,{x_{{2}}}^{3}+{x_{{3}}}^{2}}{x_0^{2}}}=C, $$ but I cant find $x_i$ as functions of $t.$ Is it possible?

Answer 1

Maple 2019.1 with "Physics" package version 419:

sol := dsolve([diff(x0(t), t) = 3*x1(t), diff(x1(t), t) = 0, diff(x2(t), t) = (2*x1(t)*x2(t) + x3(t))/x0(t), diff(x3(t), t) = (3*x1(t)*x3(t) - 6*x2(t)^2)/x0(t)])

I have:

[{x1(t) = _C2}, {x0(t) = Int(3*x1(t), t) + _C1}, {diff(x2(t), t, t) = (-6*x1(t)^2*x2(t) + 2*x1(t)*diff(x2(t), t)*x0(t) - 6*x2(t)^2)/x0(t)^2}, {x3(t) = -2*x1(t)*x2(t) + diff(x2(t), t)*x0(t)}]

If I add build command to dsolve I have a solution:

{x0(t) = 3*_C2*t + _C1, x1(t) = _C2, x2(t) = -9*(_C2*t + _C1/3)^2*WeierstrassP(((3*_C2*t + _C1)^(1/3) + _C3*_C2)/_C2, 0, _C4)/(3*_C2*t + _C1)^(4/3), x3(t) = -(3*_C2*t + _C1)*WeierstrassPPrime(((3*_C2*t + _C1)^(1/3) + _C3*_C2)/_C2, 0, _C4)}

In LaTex code:

$\left\{ {\it x0} \left( t \right) =3\,{\it c2}\,t+{\it c1},{\it x1} \left( t \right) ={\it c2},{\it x2} \left( t \right) =-9\,{\frac { \left( {\it c2}\,t+{\it c1}/3 \right) ^{2}}{ \left( 3\,{\it c2}\,t+{ \it c1} \right) ^{4/3}}{\it WeierstrassP} \left( {\frac {\sqrt [3]{3\, {\it c2}\,t+{\it c1}}+{\it c3}\,{\it c2}}{{\it c2}}},0,{\it c4} \right) },{\it x3} \left( t \right) =-{\it WeierstrassPPrime} \left( {\frac {\sqrt [3]{3\,{\it c2}\,t+{\it c1}}+{\it c3}\,{\it c2}}{{\it c2 }}},0,{\it c4} \right) \left( 3\,{\it c2}\,t+{\it c1} \right) \right\} $

Check by odetest:

odetest(sol, [diff(x0(t), t) = 3*x1(t), diff(x1(t), t) = 0, diff(x2(t), t) = (2*x1(t)*x2(t) + x3(t))/x0(t), diff(x3(t), t) = (3*x1(t)*x3(t) - 6*x2(t)^2)/x0(t)])

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