# Solution of an Ordinary Differential Equation

I'd like to know if there is an analytical solution of the following Ordinary Differential Equation :

$\displaystyle{\dfrac{d}{dt}x_i(t) = D_i + \sum_{j=1}^{n}L_{ij}x_j(t) + \sum_{j=1}^{n}\sum_{k=1}^{n}C_{ijk}x_j(t)x_k(t)}$

where $\forall i,j,k=1,...,n$ we have $x_i(t) \in M_{T\times 1}(\mathbb{R})$ and $D_i,L_{ij},C_{ijk} \in \mathbb{R}$

The linear case :

$\displaystyle{\dfrac{d}{dt}x_i(t) = D_i + \sum_{j=1}^{n}L_{ij}x_j(t)}$

is straightforward, but i don't see how to solve this ODE with the quadratic terms.

• Also posted to m.se, math.stackexchange.com/questions/694444/… – Gerry Myerson Feb 28 '14 at 22:34
• While Alexandre's discussion and example below are important ones to have, one should bear in mind that the answer to your question really does depend on the constants $C$, $L$, and $D$. If the vector field $$X = \left(D^i + L^i_jx^j+C^i_{jk} x^j x^k\right)\frac{\partial}{\partial x^i}$$ happens to lie in a finite dimensional Lie algebra of a known transitive action of a Lie group on some completion of $\mathbb{R}^n$ (as it always does when $C=0$), then there will be an explicit elementary solution of the above ODE. – Robert Bryant May 4 '15 at 8:35

In the common usage there is a word "analytic", which means "can be expanded in a convergent power series". Then there is a general theorem: if you have a differential equation $y'=F(y)$, where $y$ is a function with values in $R^n$, and $F$ is an analytic function (in your case, $F$ is a polynomial of degree $2$), then for every $t_0$ and every $y_0$, there exists a solution with the property $y(t_0)=y_0$, which is analytic in some neighborhood of $t_0$. The neighborhood depends on $F$ and on $y_0$.
Example: $$y_1'=y_1^2+y_2,\quad y_2^\prime=1.$$ This is equivalent to the single equation $y'=y^2+t+C,$ and J. Liouville proved in XIX century that it has no solutions of the type described above.
• There are, however, solutions of Alexandre's example in terms of Airy functions and their derivatives: $$y \left( t \right) ={\frac {{{\rm Ai}^{(1)}\left(-C-t\right)}{k }+{{\rm Bi}^{(1)}\left(-C-t\right)}}{{{\rm Ai}\left(-C-t\right)}{\it k}+{{\rm Bi}\left(-C-t\right)}}}$$ where $k$ is an arbitrary constant. – Robert Israel May 4 '15 at 6:59