# Solution to non-autonomous system of ODEs

I have the following system of ordinary differential equations with $\gamma(t),M(t),$ and $\beta(t)$ strictly positive and smooth.

$\dot{g_1}(t) = \gamma(t)M(t){g_2}(t)$
$\dot{g_2}(t) = \beta(t)\frac{1}{M(t)}{g_1}(t)$

with initial conditions

${g_1}(0)=1$
${g_2}(0)=1$

$0\le t \le 1$

How would I go about finding a non-trivial solution?

Editing to add that $\gamma(t), M(t),$ and $\beta(t)$ are finite on $[0,1]$

2nd edit: Corrected typo in domain of $t$

• I'm having a little trouble figuring out how "$0 < t < 1$" fits into your question. Is it the range of $t$ over which you seek a solution? – Robert Lewis Nov 3 '14 at 23:16
• Yes, that is correct. It is the domain on which the functions are defined and should be solved. – ELC Nov 3 '14 at 23:17
• If $M(t)$ is bounded away from zero, why is the problem considered in these terms, rather than absorbing $M(t), 1/M(t)$ into the factors $\gamma(t),\beta(t)$ respectively? – hardmath Nov 14 '14 at 12:19
• The problem originated from a biological system in which $\gamma(t), M(t),$ and $\beta(t)$ have specific meaning, which is why I prefer to keep the system in its current form instead of absorbing $M(t)$. – ELC Nov 15 '14 at 19:15

EEDDIITT: given a system $$f_1' = g f_2, \; \; \; f_2' = h f_2,$$ you get separate equations $$f_1'' = \left( \frac{g'}{g} \right) f_1' + g h f_1$$ $$f_2'' = \left( \frac{h'}{h} \right) f_2' + g h f_2.$$
There is one circumstance with a solution, a fairly restrictive one. Let $$A(t) = \int_0^t \gamma(s) M(s) ds,$$ $$B(t) = \int_0^t \frac{\beta(s)}{ M(s)} ds.$$ The restrictive condition is $$\frac{A}{B}$$ constant. In that case, the fundamental solution matrix is $$\left( \begin{array}{cc} \cosh \sqrt {AB} & \frac{\sqrt A \sinh \sqrt {AB}}{\sqrt B} \\ \frac{\sqrt B \sinh \sqrt {AB}}{\sqrt A} & \cosh \sqrt {AB} \end{array} \right).$$ With your initial condition, $$g_1(t) = \cosh \sqrt {AB} + \frac{\sqrt A \sinh \sqrt {AB}}{\sqrt B},$$ $$g_2(t) = \cosh \sqrt {AB} + \frac{\sqrt B \sinh \sqrt {AB}}{\sqrt A}.$$
The theorem used here is that, for a square matrix $H(t),$ if $H$ and $\dot{H}$ commute, then $$\frac{d}{dt} e^H = e^H \dot{H } = \dot{H} e^H.$$