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Basicaly, I had following equation to solve: $$\phi'(t) = \frac{a}{b\cdot\cos(\phi(t)) + c}$$

After integrating it, I get the following one: $$b\cdot\sin(\phi(t)) + c\cdot\phi(t) = a\cdot t$$ And that's where I spent hours. The only way for me is to try to guess the result, maybe I'm lack of some math techniques to solve it.

Will be glad for any advises and solutions.

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    $\begingroup$ You did forget the constant(s) of integration, which can be solved for with a boundary value problem. $\endgroup$ – Kwin van der Veen Sep 26 '15 at 2:11
  • $\begingroup$ @fibonatic I didn't forget it, it equeals to sero as phi(0) = 0 (and I implicitly used it). Didn't understand your last part about boundary value problem. $\endgroup$ – Alex Velickiy Sep 26 '15 at 12:01
  • $\begingroup$ A boundary value problem consists of any moment in time, lets call it time $\tau$, at a which $\phi(\tau)=\phi_\tau$. If $\tau=0$ then it is called an initial value problem, like in your case. In general for higher order differential equations you will need more constraints, but those constraints can be placed at different moments in time for boundary value problems. $\endgroup$ – Kwin van der Veen Sep 26 '15 at 12:23
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This is the result. There is no way to get $\varphi(t)$ as a combination of elementary functions.

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  • $\begingroup$ What is the reason of impossibility to get it as a combination of elementary functions? A theorem? $\endgroup$ – Alex Velickiy Sep 25 '15 at 22:15
  • $\begingroup$ @AlexVelickiy You may take a look here. $\endgroup$ – thanasissdr Sep 25 '15 at 22:55
  • $\begingroup$ @thanasissdr according to this theorem, phi(t) "in principle (though not necessarily with an analytic expression) can exist" "at least in some disk". That doesn't help. $\endgroup$ – Alex Velickiy Sep 26 '15 at 12:29

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