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I had this seperable ODE with the initial value to solve:

$$x'(t)=t(1-(x(t)^5), x(-3)=0$$

I was asked to find the maximal possible interval for the solution and I got that it is: $$(-\infty, -2.619)$$ But I need to know whether this is the right answer. The solution is a bit complicated so I prefer not to write it here So I would be happy if someone could tell if this right or wrong or if there is a website I can check in.

Thank you!

FAF
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    Show with qualitative arguments that the constant solution $x=1$ is globally stable when going backwards to $-\infty$. Explore the dynamical blow-up towards $-\infty$ when going forwards, for $|x|$ large enough $x'\approx tx^5$ which is separable. If you do numerical solutions, you can use this approximation to find increasingly better estimates for the pole location, see https://math.stackexchange.com/a/2623760/115115, https://math.stackexchange.com/a/3531744/115115 – Lutz Lehmann Jul 20 '21 at 08:40
  • @LutzLehmann. The problem is even pleasant just switching variables. Cheers :-) – Claude Leibovici Jul 20 '21 at 09:23
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    @ClaudeLeibovici : Simple separation-of-variables results in the same integral equation, $$t_{pole}^2-(-3)^2=-2\int_0^\infty\frac{du}{u^5+1}=-\frac25\Gamma(\frac15)\Gamma(\frac45)~~~~(u=-x).$$ This gives indeed then $t_{pole}=-2.6195574694533446$ – Lutz Lehmann Jul 20 '21 at 09:42
  • @LutzLehmann. Good to have another solution. Thanks for giving it. – Claude Leibovici Jul 20 '21 at 09:44

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The problem is not too bad if we switch variables making now the equation to be $$ t(x) \,t'(x)=\frac 1{1-x^5}\qquad \text{with}\qquad t(0)=-3$$ Now, use partial fraction decomposition since $$1-x^5=-(x-1)\left(x^2+\frac12 (\sqrt 5+1)x+1\right)\left(x^2-\frac12 (\sqrt 5-1)x+1\right)$$

This makes the problem not very difficult and $$\lim_{x\to -\infty } \, t(x)=-\frac{1}{5} \sqrt{225-2 \sqrt{10 \left(5+\sqrt{5}\right)} \pi }$$ which is your number.

Edit

Making it more general, if $t(0)=-k$, with $k>0$ $$\lim_{x\to -\infty } \, t(x)=-\sqrt{k^2-\frac{2}{5} \sqrt{2+\frac{2}{\sqrt{5}}} \pi }$$

Claude Leibovici
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