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I got into a problem while deriving a solution for my system.

I've ended up with equation $$\exp(3t) - \exp(2t) = A$$, where $A$ is some constant. I'm fine with numeric solution but I kinda can't figure out a way to solve this analytically, so I'd like to humbly ask for help or some advice, if there is any?

Thank you very much.

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    $\begingroup$ Analytically, it's a cubic equation for $exp(t)$ which can be solved using Cardano's formula. $\endgroup$ – Peter Franek May 18 '15 at 10:25
  • $\begingroup$ HINT: make change of variables $$x = \exp(t)$$ and solve the resulting polynomial for zeros $\endgroup$ – Vlad May 18 '15 at 10:27
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$$e^{3t}-e^{2t}=(e^{t})^3-(e^t)^2=A$$ Now set $z=e^t$ $$z^3-z^2=A$$ $$z^3-z^2-A=0$$ at which you solve for A using one of the equations for cubical solutions and then you have $$t=\ln z$$

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  • $\begingroup$ Yeah, thanks I'm gonna beat my head against a wall, that I didn't realize this sooner.. :-), Again, thanks. $\endgroup$ – Petr May 18 '15 at 10:34
  • $\begingroup$ No worries man! :) It's easy to get blinded in something $\endgroup$ – Zelos Malum May 18 '15 at 10:34
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Setting $x=e^t$, the equation turns to

$$x^3-x^2=A,$$ with $x>0$.

Such a cubic equation can be solved using the Cardano's formulas.

As you can check from the plot of the function $x^3-x^2$, there are two solutions in range $[0,1]$ (i.e. $t<0$) for $-\dfrac4{27}<A<0$ and one in range $[1,\infty]$ (i.e. $t>0$) for $A>0$.

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  • $\begingroup$ Crossed with Zelos and Vlad. $\endgroup$ – Yves Daoust May 18 '15 at 10:28

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