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I'm trying to find an analytical solution to the following equation:

$$a_1 x''(t)+a_2x'(t)+a_3x(t)^\alpha+a_4t+a_5=0$$

where $0<\alpha<1$. Any suggestions would be greatly appreciated.

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    $\begingroup$ It's unlikely to have closed-form general solutions. $\endgroup$ – Robert Israel Feb 6 at 18:51
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I can give you some solutions in the case $\alpha = 1/2$. Namely, in this case $x(t) = (b_1 t + b_0)^2$ is a solution when $b_1$ is a root of the quadratic $$ 2 a_2 b_1^2 + a_3 b_1 + a_4 = 0$$ and $$b_0 = -{\frac {2\,{a_1}\,{{ b_1}}^{2}+{ a_5}}{2\,{ a_2}\,{ b_1}+{ a_3}}}$$ and $b_1 t + b_0 > 0$.

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  • $\begingroup$ Thank you. Do you think of there are other solutions in the case $\alpha=\frac{1}{2}$? $\endgroup$ – user64735 Feb 6 at 19:41
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    $\begingroup$ Yes, of course there are other solutions by the Existence and Uniqueness Theorem. But there might not be other closed-form solutions. $\endgroup$ – Robert Israel Feb 6 at 20:17
  • $\begingroup$ Thanks. Would there be some value of $0<\alpha<1$ for which closed-formed general solutions can be obtained? $\endgroup$ – user64735 Feb 6 at 20:47
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    $\begingroup$ I doubt it very much. $\endgroup$ – Robert Israel Feb 6 at 21:12
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    $\begingroup$ Of course you can't do that in general. What if $f = x$? What if $f = x + 1$? $\endgroup$ – Robert Israel Feb 7 at 1:48

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