1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – J.G. Dec 31 '17 at 11:19
  • $\begingroup$ @J.G. It's a typo, sorry. $\endgroup$ – Alure Dec 31 '17 at 11:21
0
$\begingroup$

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 $x_2=\exp (\frac{5}{2}\sinh 2t)(x_{20}+\frac{x_{10}}{4}(1-\exp(-2\sinh 2t)))$.

$\endgroup$
  • $\begingroup$ J.G. What is $x_{10}$? $\endgroup$ – Alure Dec 31 '17 at 12:00
  • $\begingroup$ The value of $x_1$ at $t=0$. Similarly with $x_{20}$. $\endgroup$ – J.G. Dec 31 '17 at 12:09
  • $\begingroup$ J.G. Sorry, you're right. Thanks! $\endgroup$ – Alure Dec 31 '17 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.