Solving time derivative of glycogen dynamics: $17.6{dG\over dt} = 2000 - 13G^2$

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Can I find G, glycogen level at time t=5, if glycogen dynamics are described by the following derivative:

$$17.6{dG\over dt} = 2000 - 13G^2$$

It's been a long time since I've messed with derivatives and integrals, so I'm not sure how to start to work this problem. If anyone can walk me through this, it would be much appreciated.

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If the differnetial equation is $$17.6\,G' = 2000 - 13G^2$$ it is better to switch variables and to consider $$\frac {17.6}{t'}=2000 - 13G^2\implies t'=\frac {17.6}{ 2000 - 13G^2}$$ which is easy to solve after partial fraction decomposition. This would give $$t+C=\frac{11}{25 \sqrt{65}}\log \left(\frac{20 \sqrt{65}+13G}{20 \sqrt{65}-13 G}\right)$$ and using the condition given in comments $$C=\frac{11}{25 \sqrt{65}}\log \left(\frac{8013+80 \sqrt{65}}{7987}\right)$$ Now, solving for $G$, the awful $$G=20 \sqrt{\frac{5}{13}}\,\frac{a e^{bt} -1} {a e^{bt} +1 }\qquad a=\left(\frac{8013+80 \sqrt{65}}{7987}\right)^{\frac{25 \sqrt{65}}{11}} \qquad b=\frac{25 \sqrt{65}}{11}$$ that could be transformed as a $\tanh(.)$ function.

Computing for $t=5$, this gives $G=12.4035$ which is almost the steady state (compare with $2000-13G^2=0 \implies G=20 \sqrt{\frac{5}{13}}$.