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I'm interested of finding a closed formula for the fundamental matrix to the system $$\eqalign{ & y'(t) = a(t)z(t) \cr & z'(t) = \delta a(t)y(t) \cr} $$ $$(y(0),z(0)) = ({y_0},{z_0})$$ where $\delta$ is some constant and $a$ is a regular function. Thank you.

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Let $A(t)=\int_0^ta(s)\,ds$. The matrices $$ \begin{pmatrix}0 & a\\ \delta\,a & 0\end{pmatrix}\text{ and } \begin{pmatrix}0 & A\\ \delta\,A & 0\end{pmatrix} $$ commute. The fundamental matrix is then $$ \exp\left(\begin{pmatrix}0 & A\\ \delta\,A & 0\end{pmatrix}\right)\ . $$ If $\delta=\mu^2>0$, this is $$ \begin{pmatrix}\cosh(\mu\,A) & \dfrac{\sinh(\mu\,A)}{\mu}\\ \mu\sinh(\mu\,A) & \cosh(\mu\,A)\end{pmatrix}\ . $$ If $\delta=-\mu^2<0$, then it is $$ \begin{pmatrix}\cos(\mu\,A) & \dfrac{\sin(\mu\,A)}{\mu}\\ -\mu\sin(\mu\,A) & \cos(\mu\,A)\end{pmatrix}\ . $$

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  • $\begingroup$ Thank you sir . $\endgroup$ – Gustave Feb 28 at 15:20
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Note, that the system has an invariant. The invariant is given by

$$\Phi(y,z) = \delta\cdot y^2-z^2=\text{const.}$$

You can use this to reduce the system to a scalar differential equation. If you have found the solution it should not be a problem to derive the fundamental matrix of this system.

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  • $\begingroup$ Thank you for replying sir. $\endgroup$ – Gustave Mar 4 at 13:25

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