Solving a differential equation system

Solve the following differential equation system:

$$\begin{cases} x_1' = x_1\cdot cosh(2\cdot t)\\ x_2' = x_1\cdot cosh(2\cdot t)+ 5\cdot cosh(2\cdot t) x_2\\ x_3' = 89x_3 + 21x_4 -13x_5 -27x_6\\ x_4' = 53x_3 + 65x_4 -25x_5 -15x_6\\ x_5' = 10x_3 + 2x_4 + 46x_5 + 2x_6\\ x_6' = 56x_3 - 8x_4 - 24x_5 + 56x_6\end{cases}$$

and find a fundamental matrix $\phi (t)$ such that $\phi(0)=Id.$

The first thing I notice is that $x_1'$ and $x_2'$ equations are not related with the others, so I can attack this problem as if they were two separated systems:

$$\begin{cases} x_1' = x_1\cdot cosh(2\cdot t)\\ x_2' = x_1\cdot cosh(2\cdot t)+ 5\cdot cosh(2\cdot t)x_2\\ \end{cases}$$

$$\begin{cases}x_3' = 89x_3 + 21x_4 -13x_5 -27x_6\\ x_4' = 53x_3 + 65x_4 -25x_5 -15x_6\\ x_5' = 10x_3 + 2x_4 + 46x_5 + 2x_6\\ x_6' = 56x_3 - 8x_4 - 24x_5 + 56x_6\end{cases}$$

right?

After that, I tried to solve the second system with sagemath. The solution is ugly, but it can find a solution. But sagemath is not able to find a solution for the first one.

Why is that? How can we solve the first one?

• The fact that the second system of equations is of the form $\mathbf{y}'=A\mathbf{y}$ with a constant square matrix $A$ might be why sagemath found it so easy. For the other equation system, we first have $\ln x_1=\int\cosh 2t dt=\frac{1}{2}\sinh 2t+C$ so $x_1=x_{10}\exp(\frac{1}{2}\sinh 2t)$. The last equation is hardest of all because of its nonlinearity, unless you have a typo. – J.G. Dec 31 '17 at 11:19
• @J.G. It's a typo, sorry. – Alure Dec 31 '17 at 11:21

As I mentioned in a comment, $x_1=x_{10}\exp(\frac{1}{2}\sinh 2t)$. The last equation is $x_2'+Px_2=Q$ with $P=-5\cosh 2t,\,Q=x_{10}\cosh 2t \exp(\frac{1}{2}\sinh 2t)$. Since $R:=\exp\int P dt=\exp (-\frac{5}{2}\sinh 2t)$ (to within a multiplicative constant), the solution is $R^{-1}\int RQ dt$. So the hard part is evaluating $\int RQ dt=\int x_{10}\cosh 2t \exp(-2\sinh 2t) dt=C+\frac{x_{10}}{4}(1-\exp(-2\sinh 2t))$, and $x_2=\exp (\frac{5}{2}\sinh 2t)(x_{20}+\frac{x_{10}}{4}(1-\exp(-2\sinh 2t)))$.
• J.G. What is $x_{10}$? – Alure Dec 31 '17 at 12:00
• The value of $x_1$ at $t=0$. Similarly with $x_{20}$. – J.G. Dec 31 '17 at 12:09